Answer:
36.37% is the percent yield of the reaction.
Explanation:
[tex]4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)[/tex]
1)0.650 L nitrogen gas , at 295 K and 1.01 bar.
Let the moles of nitrogen gas be n.
Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)
Temperature of the gas = T = 295 K
Volume of the gas = V = 0.650 L
Using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol[/tex]
2) Moles of ammonia gas=[tex]\frac{2.53 g}{17 g/mol}=0.1488 mol[/tex]
Moles of oxygen gas =[tex]\frac{3.53 g}{32 g/mol}=0.1101 mol[/tex]
According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.
Then,0.1101 mol of oxygen will react with:
[tex]\frac{4}{3}\times 0.1101 mol=0.1468 mol[/tex] of ammonia.
Hence, oxygen gas is in limiting amount and act as limiting reagent.
3) Theoretical yield of nitrogen gas :
According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.
Then 0.1101 mol of oxygen will give:
[tex]\frac{2}{3}\times 0.1101 mol=0.0734 mol[/tex] of nitrogen.
Theoretical yield of nitrogen gas = 0.0734 mol
Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol
Percentage yield:
[tex]\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100[/tex]
Percentage yield of the reaction:
[tex]\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%[/tex]
36.37% is the percent yield of the reaction.
Monosodium glutamate (MSG), a flavor enhancer, has been blamed for "Chinese restaurant syndrome," the symptoms of which are headaches and chest pains. MSG has the following composition by mass: 35.51 percent C, 4.77 percent H, 37.85 percent O, 8.29 percent N, and 13.60 percent Na. What is its molecular formula if its molar mass is about 169 g?
Answer:
The molecular formula of mono sodium glutamate is [tex]C_5H_8O_4N_1Na_1[/tex]
Explanation:
Molar mass of sodium glutamate,M = 169 g/mol
let the molecular formula be [tex]C_aH_bO_cN_dNa_e[/tex]
Percentage of carbon in the M.S.G. =35.52 %
[tex]35.51\%=\frac{a\times 12 g/mol}{169 g/mol}[/tex]
a = 5
Percentage of Hydrogen in the M.S.G. = 4.77 %
[tex]4.77\%=\frac{b\times 1 g/mol}{169 g/mol}[/tex]
b = 8
Percentage of oxygen in the M.S.G. =37.85 %
[tex]8.29\%=\frac{c\times 16 g/mol}{169 g/mol}[/tex]
c = 3.99 ≈ 4
Percentage of nitrogen in the M.S.G. = 8.29 %
[tex]4.77\%=\frac{d\times 14 g/mol}{169 g/mol}[/tex]
d = 1
Percentage of sodium in the M.S.G. =13.60 %
[tex]13.60\%=\frac{c\times 23g/mol}{169 g/mol}[/tex]
e = 0.99 ≈ 1
The molecular formula be :[tex]C_aH_bO_cN_dNa_e=C_5H_8O_4N_1Na_1[/tex]
The molecular formula of Monosodium glutamate is C2H4O4N2Na2
To determine the molecular formula of monosodium glutamate based on its elemental composition, we'll first find the empirical formula, and then calculate the molecular formula.
Find the moles of each element:
Carbon (C): 35.51%
Hydrogen (H): 4.77%
Oxygen (O): 37.85%
Nitrogen (N): 8.29%
Sodium (Na): 13.60%
Calculate the moles of each element using their molar masses:
Moles of C = (35.51/100) * 169 g / (12.01 g/mol) ≈ 5.97 moles
Moles of H = (4.77/100) * 169 g / (1.01 g/mol) ≈ 7.90 moles
Moles of O = (37.85/100) * 169 g / (16.00 g/mol) ≈ 8.37 moles
Moles of N = (8.29/100) * 169 g / (14.01 g/mol) ≈ 9.99 moles
Moles of Na = (13.60/100) * 169 g / (22.99 g/mol) ≈ 9.43 moles
Find the smallest whole number ratio of moles.
Divide all moles by the smallest number of moles (approximately 5.97).
Empirical Formula:
C1H1.32O1.4N1.68Na1.58
Round the subscripts to whole numbers (since you can't have fractions of atoms):
CH2O2N2Na2
The empirical formula of Monosodium glutamate is CH2O2N2Na2.
To find the molecular formula, you need to determine the molar mass of the empirical formula and compare it to the given molar mass (169 g/mol).
The empirical formula mass is 85 g/mol (approximately), which is half of the molar mass.
Therefore, the molecular formula is twice the empirical formula:
Molecular Formula: C2H4O4N2Na2
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The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate ΔG for this reaction.
*** Please explain the reactions since I’m very confused as to wich side I should put the electrons.
Ex: Cu-> Cu2+ + 2e
Answer : The [tex]\Delta G[/tex] for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,
[tex]Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)[/tex]
Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half oxidation-reduction reaction will be :
Oxidation : [tex]Cu\rightarrow Cu^{2+}+2e^-[/tex]
Reduction : [tex]2Ag^++2e^-\rightarrow 2Ag[/tex]
Now we have to calculate the Gibbs free energy.
Formula used :
[tex]\Delta G^o=-nFE^o[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
[tex]E^o[/tex] = standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.
[tex]\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole[/tex]
Therefore, the [tex]\Delta G[/tex] for this reaction is, -88780 J/mole.
"The change in Gibbs free energy ΔG for the reaction is -94.48 kJ/mol.
To calculate ΔG for the reaction involving the reduction of silver ions with elemental copper, we can use the standard cell potential E° and the following relationship:
[tex]\[ \Delta G = -nFE\° \][/tex]
where:
- [tex]\( \Delta G \)[/tex] is the change in Gibbs free energy in joules (J).
- n is the number of moles of electrons transferred in the reaction.
- F is the Faraday constant, which is approximately 96485 J/(V·mol).
- E° is the standard cell potential in volts (V).
The balanced chemical equation for the reaction is:
[tex]\[ \text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s) \][/tex]
From the equation, we can see that two moles of electrons (n = 2) are transferred when one mole of copper is oxidized to form one mole of copper(II) ions.
Given that the standard cell potential E° is 0.46 V, we can now plug in the values:
[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[ \Delta G = -(2 \text{ mol})(96485 \text{ J/(V\·mol)})(0.46 \text{ V}) \] \[ \Delta G = -(2)(96485)(0.46) \] \[ \Delta G = -186177 \text{ J} \][/tex]
To convert joules to kilojoules, we divide by 1000:
[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[ \Delta G = -\frac{186177}{1000} \text{ kJ} \] \[ \Delta G = -186.177 \text{ kJ/mol} \] \[ \Delta G \approx -186 \text{ kJ/mol} \][/tex]
However, there seems to be a discrepancy in the significant figures used in the calculation. The standard cell potential was given to two decimal places (0.46 V), so the final answer should be rounded accordingly:
Upon re-evaluating the calculation with the correct rounding:
[tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \[ \Delta G = -(2)(96485)(0.46) \] \[ \Delta G = -94480 \text{ J/mol} \] \[ \Delta G = -94.48 \text{ kJ/mol} \][/tex]
Therefore, the correct change in Gibbs free energy for the reaction is -94.48 kJ/mol."
Consider the following equilibrium system involving SO2, Cl2, and SO2Cl2 (sulfuryl dichloride): SO2(g) + Cl2(g) ⇌ SO2Cl2(g) Predict how the equilibrium position would change if the temperature remains constant: a. Cl2 gas were added to the system. b. SO2Cl2 were removed from the system. c. SO2 were removed from the system.
Answer:
(a) forward direction
(b) forward direction
(c) backward direction.
Explanation:
Given , the chemical reaction in equilibrium is,
SO₂(g) + Cl₂(g) ⇄ SO₂Cl₂ (g)
The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,
It states that ,
When a reaction is at equlibrium , Changing the concentration , pressure, temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the changes.
(a)
For the reaction , Cl₂ is added to the system , i.e. , increasing the concentration of Cl₂ ,Now, according to Le Chatelier , The reaction will move in forward direction , to reduce the increased amount of Cl₂.
Hence, reaction will go in forward direction.
(b)
Removing SO₂Cl₂ from the system ,i.e. , decreasing the concentration of SO₂Cl₂ , according to Le Chatelier , the reaction will move in forward direction , to increase the amount of reduced SO₂Cl₂.
Hence, reaction will go in forward direction.
(c)
Removing SO₂ from the system , i.e. decreasing the concentration of SO₂ , according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced SO₂.
Hence, reaction will go in backward direction.
Glucose prefers an open-chain conformation in aqueous solution(T/F)
Answer:
False
Explanation:
Glucose is a monosachharide carbohydrate, with the molecular formula C₆H₁₂O₆.
Glucose molecule can exist in two forms-
1. Open chain form
2. cyclic form
The open chain form of the glucose is an unbranched 6 carbon atom chain. The carbon 1 of the molecule is an aldehyde group and the rest of the five carbon atoms have one hydroxyl group each.
The cyclic form of the glucose can be-
a. Pyranose: The pyranose form is a 6-membered cyclic ring, which consists of 5 carbon atoms and 1 oxygen atom in the ring.
b. Furanose: The furanose form is a 5- membered cyclic ring, which consists of 4 carbon atoms and 1 oxygen atom in the ring.
In an aqueous solution, 99% glucose molecule exists in the cyclic pyranose form as it is energetically more stable.
Therefore, in aqueous solution, the glucose molecule does not prefer the open-chain conformation.
Therefore, the statement is false.
Consider the following reaction in chemical equilibrium: 2BrNO(g)⇌2NO(g)+Br2(g) What is the effect of adding additional Br2(g) to the reaction mixture? What is the effect of adding additional BrNO? Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
Answer:The reaction would shift to the left on adding Br2(g)
The reaction would shift to right on adding BrNO(g)
Explanation:
The concept can be explained on the basis of LeChateliers principle.
LeChateliers principle explain the effect on equilibrium when a reaction system at equilibrium is subjected to any disturbance or change in conditions then the reaction shifts in such a way so as to reduce the effect of that change or disturbance and again establish the equilibrium.
For example if we add more reactants in a given reaction at equilibrium, the reaction would shift in such a way so that it can reduce the effect of increasing the concentration of reactants and hence the reaction would favor that direction in which it can reduce the concentration of reactants.So when we increase the concentration of reactants the reaction would move towards the formation of more products and so the concentration of reactants would be less in the reaction.
Likewise if we increase the concentration of products so the reaction would shift in such a way so that it can oppose the increased concentration of products so the reaction moves towards more formation of reactants that is the products decompose to form reactants and reaction moves backwards.
In this reaction:
2BrNO(g)⇄2NO(g)+Br₂(g)
When we add more amount of Br₂(g) the reaction would proceed in such a ways so that oppose the increased concentration of Br₂(g).Hence the reaction would move towards left that is backwards.
When we add more amount of BrNO(g)the reaction would proceed in such a ways so that oppose the increased concentration of BrNO(g).Hence the reaction would move towards right that is forward.
For each of the following acid-base reactions, calculate how many grams of each acid are necessary to completely react with and neutralize 2.7 g of the base.Part A : HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)Part B : 2HNO3(aq)+Ca(OH)2(aq)→2H2O(l)+Ca(NO3)2(aq)Part C : H2SO4(aq)+2KOH(aq)→2H2O(l)+K2SO4(aq)
Explanation:
A) [tex]HCl(aq)+NaOH(aq)\rightarrow H_2O(l)+NaCl(aq)[/tex]
Mass of sodium hydroxide= 2.7 g
Moles of base = [tex]\frac{2.7 g}{40 g/mol}=0.0675 mol[/tex]
According to reaction , 1 mol of NaOH neutralizes with 1 mol of HCl.
Then 0.0675 mol of base will neutralize:
[tex]\frac{1}{1}\times 0.0675 mol=0.0675 mol[/tex] of HCl.
Mass of 0.0675 mol of HCl = 0.0675 mol × 35.5 g/mol = 2.396 g
2.396 grams of acid will completely react with and neutralize 2.7 g of the sodium hydroxide.
B) [tex]2HNO_3(aq)+Ca(OH)_2(aq)\rightarrow 2H_2O(l)+Ca(NO_3)_2(aq)[/tex]
Mass of calcium hydroxide= 2.7 g
Moles of base = [tex]\frac{2.7 g}{57 g/mol}=0.04736 mol[/tex]
According to reaction , 2 mol of [tex]HNO_3[/tex] neutralizes with 1 mol of [tex]Ca(OH)_2[/tex].
Then 0.04736 mol of base will neutralize:
[tex]\frac{2}{1}\times 0.04736 mol=0.09472 mol[/tex] of [tex]HNO_3[/tex]
Mass of 0.09472 mol of [tex]HNO_3[/tex] :
0.09472 mol × 63g/mol = 5.9673 g
5.9673 grams of acid will completely react with and neutralize 2.7 g of the calcium hydroxide.
C) [tex]H_2SO_4(aq)+2KOH(aq)\rightarrow 2H_2O(l)+K_2SO_4(aq)[/tex]
Mass of potassium hydroxide= 2.7 g
Moles of base = [tex]\frac{2.7 g}{56 g/mol}=0.04821 mol[/tex]
According to reaction , 1 mol of [tex]H_2SO_4[/tex] neutralizes with 2 mol of [tex]KOH[/tex].
Then 0.04821 mol of base will neutralize:
[tex]\frac{1}{2}\times 0.04821 mol=0.02410 mol[/tex] of [tex]H_2SO_4[/tex]
Mass of 0.02410 mol of [tex]H_2SO_4[/tex] :
0.02410 mol × 98 g/mol = 2.3618 g
2.3618 grams of acid will completely react with and neutralize 2.7 g of the potassium hydroxide.
The amount, in grams, of each acid that would be needed for each of the reactions represented by the equations respectively, would be 2.46 grams, 4.59 grams, and 2.36 grams.
Stoichiometric calculationsFrom the first equation: HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)
The mole ratio of HCl to NaOH is 1:1.
Mole of 2.7 g NaOH = 2.7/40 = 0.0675 moles
Equivalent mole of HCl = 0.0675 moles
Mass of 0.0675 mole HCl = 0.0675 x 36.458 = 2.46 grams
For the second equation: 2HNO3(aq)+Ca(OH)2(aq)→2H2O(l)+Ca(NO3)2(aq)
Mole ratio of base to acid = 1:2
Mole of 2.7 grams Ca(OH)2 = 2.7/74.093 = 0.0364 moles
Equivalent mole of HNO3 = 0.0364 x 2 = 0.0728 moles
Mass of 0.0728 mole HNO3 = 0.0728 x 63.01 = 4.59 grams
For the third equation: H2SO4(aq)+2KOH(aq)→2H2O(l)+K2SO4(aq)
Mole ratio of acid to base = 1;2
Mole of 2.7 grams KOH = 2.7/56.1 = 0.0481 moles
Equivalent mole of H2SO4 = 0.0481/2 = 0.024 moles
Mass of 0.024 mole H2SO4 = 0.024 x 98.079 = 2.36 grams
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Copper crystallizes with a face-centered cubic lattice and has a density of 8.93 g/cm3.
a.) Calculate the mass of one unit cell of copper (in grams) b.) Calculate the volume of the copper unit cell (in cm3). c.) Calculate the edge length of the unit cell (in cm). d.) Calculate the radius of a copper atom (in pm).
Answer:
For a: The mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]
For b: The volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]
For c: The edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]
For d: The radius of a copper atom 127.82 pm.
Explanation:
For a:We know that:
Mass of copper atom = 63.55 g/mol
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.
If, [tex]6.022\times 10^{23}[/tex] number of atoms occupies 63.55 grams.
So, 1 atom will occupy = [tex]\frac{63.55g}{6.022\times 10^{23}atom}\times 1 atom=1.0553\times 10^{-22}g[/tex]
Hence, the mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]
For b:Copper crystallizes with a face-centered cubic lattice. This means that 4 number of copper atoms are present in 1 units cell.
Mass of 4 atoms of copper atom = [tex]1.0553\times 10^{-22}g/atom \times 4atoms=4.2212\times 10^{-22}g[/tex]
We are given:
Density of copper = [tex]8.93g/cm^3[/tex]
To find the volume of copper, we use the equation:
[tex]\text{Density of copper}=\frac{\text{Mass of copper}}{\text{Volume of copper}}[/tex]
Putting values in above equation, we get:
[tex]8.93g/cm^3=\frac{4.2212\times 10^{-22}}{\text{Volume of copper}}\\\\\text{Volume of copper}=4.726\times 10^{-23}cm^3[/tex]
Hence, the volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]
For c:Edge length of the unit cell is taken as 'a'
Volume of cube = [tex]a^3[/tex]
Putting the value of volume of unit in above equation, we get:
[tex]\sqrt[3]{4.726\times 10^{-23}}cm^3=3.615\times 10^{-8}cm[/tex]
Hence, the edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]
For d:The relation of radius and edge length for a face-centered lattice follows:
[tex]a=r\sqrt{8}[/tex]
Putting values in above equation, we get:
[tex]3.615\times 10^{-8}=r\sqrt{8}\\\\r=1.2782\times 10^{-8}cm[/tex]
Converting cm to pm, we get:
[tex]1cm=10^{10}pm[/tex]
So, [tex]1.2782\times 10^{-8}cm=127.82pm[/tex]
Hence, the radius of a copper atom 127.82 pm.
The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 6.200 g C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water?
Answer:
31.9178 °C is the final temperature of the water
Explanation:
[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ[/tex]
Mass of benzene burned = 6.200 g
Moles of benzene burned = [tex]\frac{6.200 g}{78 g/mol}=0.0794 mol[/tex]
According to reaction , 2 moles of benzene gives 6542 kJ of energy on combustion.
Then 0.0794 mol of benzene on combustion will give:
[tex]\frac{6542 kJ}{2}\times 0.0794 kJ=259.7174 kJ=Q[/tex]
Mass of water in which Q heat is added = m = 5691 g
Initial temperature = [tex]T_i=21^oC[/tex]
Final temperature = [tex]T_f[/tex]
Specific heat of water = c = 4.18 J/g°C
Change in temperature of water = [tex]T_f-T_i[/tex]
[tex]Q=mc\Delta t=mc(T_f-T_i)[/tex]
[tex]259,717.4 J=5691 g\times 4.18 J/g^oC\times (T_f-21^oC)[/tex]
[tex]T_f=31.91 ^oC[/tex]
31.9178 °C is the final temperature of the water
The final temperature of the water : 31.916 °C
Further explanationThe law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = mc∆TQ = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
From reaction:
2C₆H₆ (l) + 15O₂ (g) ⟶12CO₂ (g) + 6H₂O (l) +6542 kJ, heat released by +6542 kJ to burn 2 moles of C₆H₆
If there are 6,200 g of C₆H₆ then the number of moles:
mol = mass: molar mass C₆H₆
mol = 6.2: 78
mol C6H6 = 0.0795
so the heat released in combustion 0.0795 mol C6H6:
[tex]\rm Q=heat=\dfrac{0.0795}{2}\times 6542\:kJ\\\\Q=260.0445\:kJ[/tex]
the heat produced from the burning is added to 5691 g of water at 21 ∘ C
So :
Q = m . c . ∆T (specific heat of water = 4,186 joules / gram ° C)
260044.5 = 5691 . 4.186.∆T
[tex]\rm \Delta T=\dfrac{260044.5}{5691\times 4.186}\\\\\Delta T=10.916\\\\\Delta T=T(final)-Ti(initial)\\\\10.916=T-21\\\\T=31.916\:C[/tex]
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The reaction of ethyl acetate with sodium hydroxide, CH3COOC2H5(aq)+NaOH(aq)⇌CH3COONa(aq)+C2H5OH(aq) is first order in CH3COOC2H5 and first order in NaOH. If the concentration of CH3COOC2H5 was increased by half and the concentration of NaOH was quadrupled, by what factor would the reaction rate increase?
Answer:
so the reaction rate increases by a factor 6.
Explanation:
For the given equation the reaction is first order with respect to both ester and sodium hydroxide
So we can say that the rate law is
[tex]Rate(initial)=K[NaOH][CH_{3}COOC_{2}H_{5}][/tex]
now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration
The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.
The new rate law is
[tex]Rate(final)=K[1.5XNaOH][4XCH_{3}COOC_{2}H_{5}][/tex]
the final rate = 6 X initial rate
so the reaction rate increases by a factor 6.
The reaction rate of ethyl acetate with sodium hydroxide would increase by a factor of 6 if the concentration of ethyl acetate is increased by half and the concentration of NaOH is quadrupled, as it is first order in both reactants.
The reaction of ethyl acetate with sodium hydroxide is:
CH₃COOC₂H₅(aq) + NaOH(aq) ⇄ CH₃COONa(aq) + C₂H₅OH(aq)This reaction is first order concerning both CH₃COOC₂H₅ and NaOH. The rate law for this reaction can be written as:
Rate = k[CH₃COOC₂H₅][NaOH]If the concentration of CH₃COOC₂H₅ is increased by half, its new concentration becomes 1.5 times its initial concentration. If the concentration of NaOH is quadrupled, its new concentration becomes 4 times its initial concentration. Therefore, the rate of the reaction increases by a factor of:
1.5 (from CH₃COOC₂H₅) x 4 (from NaOH) = 6So, the reaction rate would increase by a factor of 6.
f 23.2 grams of an aqueous solution of nickel(II) iodide, NiI2, contains 5.47 grams of nickel(II) iodide, what is the percentage by mass of nickel(II) iodide in the solution?
Answer:
the percentage by mass of Nickel(II) iodide = 23.58%
Explanation:
% by mass of solute = (mass of solute/mass of solution) x 100%
% by mass of NiI2 = (mass of NiI2/mass of solution) x 100%
% by mass of NiI2 = (5.47 grams/23.2 grams) x 100% = 23.58% m/m
Diethyl ether is produced from ethanol according to the following equation: 2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l) Calculate the percent yield if 65.2 g of ethanol reacts to produce 17.2 g of ether.
Answer : The percent yield is, 32.79 %
Explanation :
First we have to calculate the moles of [tex]CH_3CH_2OH[/tex].
[tex]\text{Moles of }CH_3CH_2OH=\frac{\text{Mass of }CH_3CH_2OH}{\text{Molar mass of }CH_3CH_2OH}=\frac{65.2g}{46.07g/mole}=1.415mole[/tex]
Now we have to calculate the moles of [tex]CH_3CH_2OCH_2CH_3[/tex]
The balanced chemical reaction will be,
[tex]2CH_3CH_2OH(l)\rightarrow CH_3CH_2OCH_2CH_3(l)+H_2O(l)[/tex]
From the balanced reaction, we conclude that
As, 2 moles of [tex]CH_3CH_2OH[/tex] react to give 1 mole of [tex]CH_3CH_2OCH_2CH_3[/tex]
So, 1.415 moles of [tex]CH_3CH_2OH[/tex] react to give [tex]\frac{1.415}{2}=0.7075[/tex] mole of [tex]CH_3CH_2OCH_2CH_3[/tex]
Now we have to calculate the mass of [tex]CH_3CH_2OCH_2CH_3[/tex]
[tex]\text{Mass of ether}=\text{Moles of ether}\times \text{Molar mass of ether}[/tex]
[tex]\text{Mass of }ether=(0.7075mole)\times (74.12g/mole)=52.44g[/tex]
The theoretical yield of ether, [tex]CH_3CH_2OCH_2CH_3[/tex] = 52.44 g
Now we have to calculate the percent yield of [tex]CH_3CH_2OCH_2CH_3[/tex]
[tex]\%\text{ yield of ether}=\frac{\text{Actual yield of ether}}{\text{Theoretical yield of ether}}\times 100=\frac{17.2g}{52.44g}\times 100=32.79\%[/tex]
Therefore, the percent yield is, 32.79 %
Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules.
Answer : The Lewis structures for the two molecules are shown below.
Solution : Given,
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 52.2 g
Mass of H = 13.1 g
Mass of O = 34.7 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{52.2g}{12g/mole}=4.35moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{13.1g}{1g/mole}=13.1moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{34.7g}{16g/mole}=2.17moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.35}{2.17}=2.00\approx 2[/tex]
For H = [tex]\frac{13.1}{2.17}=6.03\approx 6[/tex]
For O = [tex]\frac{2.17}{2.17}=1[/tex]
The ratio of C : H : O = 2 : 6 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_2H_6O_1[/tex] = [tex]C_2H_6O[/tex]
The empirical formula weight = 12(2) + 6(1) + 1(16) = 46 gram/eq
Now we have to calculate the value of 'n'.
Formula used :
[tex]n=\frac{\text{Molecular formula weight}}{\text{Empirical formula weight}}=\frac{45g/mole}{46g/eq}=0.9\approx 1[/tex]
Molecular formula = [tex](C_2H_6O)_n=(C_2H_6O)_1=C_2H_6O[/tex]
So, there are two possibilities for the arrangements of atoms. That means, it will be an ethanol [tex](H_3C-CH_2-OH)[/tex] or dimethyl ether [tex](H_3C-O-CH_3)[/tex].
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
As we know that carbon has '4' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electron.
Therefore, the total number of valence electrons in [tex]C_2H_6O[/tex] = 2(4) + 6(1) + 6 = 20
According to Lewis-dot structure, there are 16 number of bonding electrons and 4 number of non-bonding electrons.
Thus, the Lewis structures for the two molecules are shown below.
We give 70 J as heat to a diatomic gas, which then expands at constant pressure. The gas molecules rotate but do not oscillate. By how much does the internal energy of the gas increase?
Answer:
50 Joule
Explanation:
Diatomic gas
Q = Heat given = 70 J
n = number of moles
Cp = specific heat at constant pressure
ΔT = Change in temperature
R = Gas constant
Change in energy
ΔE = Q-w
⇒ΔE = n(Cp)ΔT-nRΔT
As it is a diatomic gas Cp = (7/2)R
Putting the value of Cp in the above equation we get
Q = (7/2)RΔT
ΔE = (5/2)RΔT
Dividing the equations we get
ΔE/Q = 5/7
⇒ΔE = (5/7)Q
⇒ΔE = (5/7)×70
⇒ΔE = 50 J
∴ The internal energy change is 50 Joule
1. Wine is approximately 12% ethanol C2H5OH by volume . Ethanol has a molar mass of 46.06g/mol and a density 0.789g/mL. How many moles of ethanol are present in a 750mL bottle of wine ?
Hey there!:
Total volume of wine = 750ml
volume ℅ of ethanol = 12 %
volume of ethanol = (12ml/100ml)*750ml = 90ml
Density of Ethanol = 0.789 g/ml
Mass of Ethanol = 0.789 g/ml × 90ml = 71.01 g
Molar mass of ethanol = 46 g/mol Nº of mole of ethanol = Mass/molar mass
=> 71.01 g /46(g/mol)= 1.5437 moles
Hope this helps!
1.541 moles of ethanol are present in a 750 mL bottle of wine.
How to find the number of moles ?Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]
What is Density ?The substance per unit volume is called Density. SI unit of density is kg/m.
It is expressed as:
Density = [tex]\frac{\text{Mass}}{\text{Volume}}[/tex]
Volume of ethanol = 12%
= [tex]\frac{12}{100}[/tex]
= 0.12
Volume of ethanol = 0.12 × 750
= 90
Density of ethanol = [tex]\frac{\text{Mass of ethanol}}{\text{Volume of ethanol}}[/tex]
0.789 g/mL = [tex]\frac{\text{Mass of ethanol}}{90}[/tex]
Mass of ethanol = 0.789 × 90
= 71.01 g
Now put the value in above formula we get
Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]
= [tex]\frac{71.01\ g}{46.06\ \text{g/mol}}[/tex]
= 1.541 mol
Thus from the above conclusion we can say that 1.541 moles of ethanol are present in a 750 mL bottle of wine.
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The reaction of hydrogen peroxide with iodine, H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) is first order in H2O2 and first order in I2. If the concentration of H2O2 was increased by half and the concentration of I2 was quadrupled, by what factor would the reaction rate increase?
Answer:So we can say that the rate of reaction increases by factor of 6.
Explanation:
The rate law for any given reaction
A+B⇄C+D
Rate law for the above reaction is:
R=K[A]ᵃ[B]ᵇ
a and b are the order of reaction and it is an experimentally determined quantity.
K is the rate constant and it is constant for a given reaction
[A] and [B] are the concentrations of the reactants.
R is the rate of reaction
For the given reaction :
H₂O₂(aq.)+I₂(aq.)⇆OH⁻(aq.)+HIO(aq.)
Also it is given for the reaction that order with respect to H₂O₂ is 1 and order with respect to I₂ is also 1
The rate law can be written as :
R=K[H₂O₂]¹[I₂]¹
k=rate constant
When we increase the concentration of H₂O₂ by half which meansthat new concentration of H₂O₂ will be= 3/2[H₂O₂].
When we increase the concentration of I₂ by 4 which means that new concentration of I₂ will be= 4[I₂]
Putting the values of our new concentration in the rate law:
R=K3/2[H₂O₂]¹4[I₂]¹
R=6K[H₂O₂]¹[I₂]¹=New rate
So as we can see that the new rate of the reaction using the new concentration is 6 times the older rate of reaction.
So we can say that the rate of reaction increases by factor of 6.
A student makes a mistake while preparing a vitamin C sample for titration and adds the potassium iodide solution twice. How will the larger quantity of KI influence the amount of NBS needed to titrate the sample? a. Increase it. b. Decrease it. c. Will have no effect on it.
The added KI does not have any impact
The reaction invovles Titration of vitaminc ( Ascorbic acid)
ascorbic acid + I₂ → 2 I⁻ + dehydroascorbic acid
the excess iodine is free reacts with the starch indicator, forming the blue-black starch-iodine complex.
This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.
Answer B
Hope this helps!
Final answer:
Adding excess potassium iodide (KI) to a vitamin C sample for titration with N-bromosuccinimide (NBS) results in increased production of iodine, which requires more sodium thiosulfate (Na₂S₂O₃) for back titration. Therefore, the amount of NBS needed to titrate the sample will increase.
Explanation:
When a student mistakenly adds a potassium iodide (KI) solution twice while preparing a vitamin C sample for titration with N-bromosuccinimide (NBS), this will lead to the addition of excess KI in the solution. The presence of additional KI will reduce the titrand more, producing a stoichiometric amount of iodine (I₂). This increased amount of I₂ will then be determined by back titration using sodium thiosulfate (Na₂S₂O₃) as a reducing titrant. Adding excess KI does not change the amount of vitamin C in the sample, but it does result in an increased production of I₂ which needs to be titrated with sodium thiosulfate. Therefore, the larger quantity of KI will increase the amount of NBS needed to titrate the sample since there is more I₂ produced that needs to be accounted for during the titration process.
Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of 6.0 grams of O2 with 7.0 grams of S. What is the % yield of SO3 in this experiment
To find the % yield of SO3, we calculate the theoretical yield based on the molar mass and stoichiometry of the reaction. The theoretical yield is shown to be 30.02 g, and with an actual yield of 7.9 g, the % yield is calculated to be 26.32%.
To calculate the % yield of SO3, we first need to determine the theoretical yield based on the reaction stoichiometry. The balanced equation is:
2SO2 + O2 → 2SO3
Given that excess sulfur (S) is present, the limiting reactant is O2. Using the molar mass of O2 (32.00 g/mol) and SO3 (80.06 g/mol), we can calculate the theoretical yield:
Calculate moles of O2: moles = mass / molar mass = 6.0 g / 32.00 g/mol = 0.1875 mol.
Stoichiometry tells us that 1 mol of O2 produces 2 mol of SO3, so the expected moles of SO3 is 2 × 0.1875 mol = 0.375 mol.
Calculate the theoretical yield of SO3: 0.375 mol × 80.06 g/mol = 30.02 g.
Next, we compare the theoretical yield to the actual yield to determine the percent yield:
Percent yield = (actual yield / theoretical yield) × 100 = (7.9 g / 30.02 g) × 100 = 26.32%
The percent yield of SO3 in this experiment is 26.32%.
The percent yield of SO3 in this experiment is 26.32%.
To find the % yield of SO3, we calculate the theoretical yield based on the molar mass and stoichiometry of the reaction. The theoretical yield is shown to be 30.02 g, and with an actual yield of 7.9 g, the % yield is calculated to be 26.32%.
To calculate the % yield of SO3, we first need to determine the theoretical yield based on the reaction stoichiometry. The balanced equation is:
2SO2 + O2 → 2SO3
Given that excess sulfur (S) is present, the limiting reactant is O2. Using the molar mass of O2 (32.00 g/mol) and SO3 (80.06 g/mol), we can calculate the theoretical yield:
Calculate moles of O2: moles = mass / molar mass = 6.0 g / 32.00 g/mol = 0.1875 mol.
Stoichiometry tells us that 1 mol of O2 produces 2 mol of SO3, so the expected moles of SO3 is 2 × 0.1875 mol = 0.375 mol.
Calculate the theoretical yield of SO3: 0.375 mol × 80.06 g/mol = 30.02 g.
Next, we compare the theoretical yield to the actual yield to determine the percent yield:
Percent yield = (actual yield / theoretical yield) × 100 = (7.9 g / 30.02 g) × 100 = 26.32%
Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2(g)+C(s)↽−−⇀CS2(g)????c=9.40 at 900 K How many grams of CS2(g) can be prepared by heating 8.08 mol S2(g) with excess carbon in a 5.35 L reaction vessel held at 900 K until equilibrium is attained?
Answer : The mass of [tex]CS_2[/tex] is, 555.028 grams
Explanation :
First er have to calculate the concentration of [tex]S_2[/tex].
[tex]\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L[/tex]
Now we have to calculate the concentration of [tex]CS_2[/tex].
The given balanced chemical reaction is,
[tex]S_2(g)+C(s)\rightleftharpoons CS_2(g)[/tex]
Initial conc. 1.51 0 0
At eqm. conc. (1.51-x) (x) (x)
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CS_2]}{[S_2]}[/tex]
Now put all the given values in this expression, we get :
[tex]9.40=\frac{x}{(1.51-x)}[/tex]
By solving the term 'x', we get :
x = 1.365 M
Concentration of [tex]CS_2[/tex] = x M = 1.365 M
Now we have to calculate the moles of [tex]CS_2[/tex].
[tex]\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole[/tex]
Now we have to calculate the mass of [tex]CS_2[/tex].
Molar mass of [tex]CS_2[/tex] = 76 g/mole
[tex]\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g[/tex]
Therefore, the mass of [tex]CS_2[/tex] is, 555.028 grams
The amount of carbon disulfide CS2 at equilibrium in the given reaction with sulfur S2 and carbon C can be determined by using the provided equilibrium constant Kc and the initial concentration of S2. The equilibrium concentration of CS2 corresponds to 14.19 M. To determine its mass in grams, this concentration is multiplied by the volume of the reaction vessel and the molar mass of CS2.
Explanation:In the given chemical reaction S2 (g) + C(s) <-> CS2 (g) in which sulfur S2 (g) and carbon C(s) combine to form carbon disulfide CS2 (g), the equilibrium constant Kc at 900K is given as 9.40.
This value of the equilibrium constant represents the ratio of the concentration of the product (CS2) to the concentration of the reactants (S2 and C). Since the reaction involves heating sulfur S2 and carbon C in excess, we can solve for the equilibrium concentration of CS2 based on the initial amount of S2 used (8.08 mol) and the volume of the reaction vessel (5.35 L).
First we calculate the initial concentration of S2 as [S2] = 8.08 mol / 5.35 L = 1.51 M. Given the reaction stoichiometry, each mol of S2 produces one mol of CS2, so at equilibrium [CS2] equals to the equilibrium concentration of S2.
Substituting these into the equilibrium expression Kc = [CS2]/[S2] and solving, we find [CS2] = Kc * [S2] = 9.40 * 1.51 M = 14.19 M, which represents the equilibrium concentration of CS2. The amount in grams of CS2 can then be found by multiplying this concentration by the volume of the vessel and the molar mass of CS2 to give the total mass of CS2 in the reaction vessel at equilibrium.
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Calculate the indicated value based on the information given: a. What is the Kb of the lactate ion? The Ka of lactic acid is 1.4 3 1024 . b. What is the Kb of the conjugate base of pyruvic acid? The Ka of pyruvic acid is 2.8 3 1023 .
Answer :
(a) The value of [tex]K_b[/tex] of the lactate ion is, [tex]7.14\times 10^{-39}[/tex]
(b) The value of [tex]K_b[/tex] of the conjugate base of pyruvic acid is, [tex]3.53\times 10^{-38}[/tex]
Explanation :
Solution for (a) :
As we are given : [tex]K_a=1.4\times 10^{24}[/tex]
As we know that,
[tex]K_a\times K_b=K_w[/tex]
where,
[tex]K_a[/tex] = dissociation constant of an acid = [tex]1.4\times 10^{24}[/tex]
[tex]K_b[/tex] = dissociation constant of a base = ?
[tex]K_w[/tex] = dissociation constant of water = [tex]1\times 10^{-14}[/tex]
Now put all the given values in the above expression, we get the dissociation constant of a base (lactate ion).
[tex]1.4\times 10^{24}\times K_b=1\times 10^{-14}[/tex]
[tex]K_b=7.14\times 10^{-39}[/tex]
Therefore, the value of [tex]K_b[/tex] of the lactate ion is, [tex]7.14\times 10^{-39}[/tex]
Solution for (b) :
As we are given : [tex]K_a=2.83\times 10^{23}[/tex]
As we know that,
[tex]K_a\times K_b=K_w[/tex]
where,
[tex]K_a[/tex] = dissociation constant of an acid = [tex]2.83\times 10^{23}[/tex]
[tex]K_b[/tex] = dissociation constant of a base = ?
[tex]K_w[/tex] = dissociation constant of water = [tex]1\times 10^{-14}[/tex]
Now put all the given values in the above expression, we get the dissociation constant of a base (conjugate base of pyruvic acid).
[tex]2.83\times 10^{23}\times K_b=1\times 10^{-14}[/tex]
[tex]K_b=3.53\times 10^{-38}[/tex]
Therefore, the value of [tex]K_b[/tex] of the conjugate base of pyruvic acid is, [tex]3.53\times 10^{-38}[/tex]
The wavelength of the red-pink line emitted by a laboratory sample of excited hydrogen is 656 nm. Taking a spectrum of a glowing nebula, you find that the same red-pink line of hydrogen appears at 662 nm. You conclude that the nebula
A. Is 1% hotter than hydrogen in the laboratory sample.
B. Is moving towards us at about 1% the speed of light.
C. Is 1% cooler than hydrogen in the laboratory sample.
D. Is moving away from us at about 1% the speed of light
Answer:
Choice D) The nebula is moving away from the observer.
Explanation:
Is the emission here a result of electron transition or thermal radiation?
The energy difference between two atomic energy levels is discrete. As a result, emissions due to electron transition exist as discrete lines.On the contrary, the thermal radiation of objects above 0 degree Kelvins exists as a continuous frequency spectrum.The red-pink emission here is as a line rather than a continuous spectrum. In other words, the red-pink line observed is a result of electron transition. The energy difference will be constant. That should be the same case on the earth as it is in space at the nebula.
Also, this energy difference does not depend on the temperature of the hydrogen. Only that at higher temperature, low-energy radiations will be less prominent. The wavelength will still be 656 nm when the light was emitted from the nebula.
The wavelength observed on the earth is longer than the wavelength emitted. The Doppler's effect is likely to be responsible. As the star moves away from the earth, the distance that light from the star needs to travel keep increasing. Consider two consecutive peaks from the star. When compared with the first peak, the second peak will need to travel a few more kilometers and will need a few more fractions of a second to get to the earth. It would appear to an observer on the earth that the frequency of the light is lower than it actually is. Accordingly, the wavelength will appear to be longer than it was when emitted from the star.
Conversely, the wavelength will appear shorter if the source is moving toward to observer. For this star, the wavelength appears to be longer than it really is. In other words, the star is moving away from the earth.
The ratio between the speed at which the star moves away from the earth and the speed of the light can be found using the equation: (Source: AstronomyOnline)
[tex]\displaystyle \frac{v}{c} = \frac{\Delta \lambda}{\lambda_0} \approx 0.009[/tex].
The gas cloud is moving away from Earth at about 1% the speed of light due to the Doppler effect.So,option D is correct.
The gas cloud is moving away from Earth at about 1% the speed of light. This can be inferred from the observed shift in the wavelength of the hydrogen emission line from 656.3 nm to 656.6 nm.
The shift in wavelength is due to the Doppler effect, indicating the motion of the source relative to the observer.
30.0 mL of 0.20 M AgNO, are added to 100.0 mL of 0.10 M HCI in a thermally nsulated vessel. The following reaction takes place: Ag (aq)+ Cl (aq)AgCI (s) The two solutions were initially at 22.00°C and the final temperature was 22.80 C Calculate the heat of this reaction in k.Jimol of AgCI formed. Assume a combined mass of 120 g and a specific heat capacity of 4.18 JK-'g for the reaction mixture.
Answer : The heat of this reaction of AgCI formed will be, 66.88 KJ
Explanation :
First we have to calculate the heat of the reaction.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = amount of heat = ?
[tex]c[/tex] = specific heat capacity = [tex]4.18J/g.K[/tex]
m = mass of substance = 120 g
[tex]T_{final}[/tex] = final temperature = [tex]22^oC=273+22=295K[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]22.8^oC=273+22.8=295.8K[/tex]
Now put all the given values in the above formula, we get:
[tex]q=120g\times 4.18J/g.K\times (295.8-295)K[/tex]
[tex]q=401.28J[/tex]
Now we have to calculate the number of moles of [tex]AgNO_3[/tex] and [tex]HCl[/tex].
[tex]\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }AgNO_3=0.20mole/L\times 0.03L=0.006mole[/tex]
[tex]\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }HCl=0.10mole/L\times 0.1L=0.01mole[/tex]
Now we have to calculate the limiting reactant.
The balanced chemical reaction will be,
[tex]AgNO_3+HCl\rightarrow AgCl+HNO_3[/tex]
As, 1 mole of [tex]AgNO_3[/tex] react with 1 mole of HCl
So, 0.006 mole of [tex]AgNO_3[/tex] react with 0.006 mole of HCl
From this we conclude that, [tex]HCl[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of AgCl.
The given balanced reaction is,
[tex]Ag^++Cl^-\rightarrow AgCl[/tex]
From this we conclude that,
1 mole of [tex]Ag^+[/tex] react with 1 mole [tex]Cl^-[/tex] to produce 1 mole of [tex]AgCl[/tex]
0.006 mole of [tex]Ag^+[/tex] react with 0.006 mole [tex]Cl^-[/tex] to produce 0.006 mole of [tex]AgCl[/tex]
Now we have to calculate the heat of this reaction of AgCI formed.
As, 0.006 mole of AgCl produced the heat = 401.28 J
So, 1 mole of AgCl produced the heat = [tex]\frac{401.28}{0.006}=66880J=66.88KJ[/tex]
Therefore, the heat of this reaction of AgCI formed will be, 66.88 KJ
Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides. (a) H2SO4 (b) Ca(OH)2 (c) BrOH (d) ClNO2 (e) TiCl4 (f) NaH'
Answer :
Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
Rules for Oxidation Numbers are :
The oxidation number of a free element is always zero.The oxidation number of a monatomic ion equals the charge of the ion.The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.The oxidation number of oxygen (O) in compounds is usually -2.The oxidation number of a Group 17 element in a binary compound is -1.The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.Now we have to determine the oxidation state of the elements in the compound.
(a) [tex]H_2SO_4[/tex]
Let the oxidation state of 'S' be, 'x'
[tex]2(+1)+x+4(-2)=0\\\\x=+6[/tex]
Hence, the oxidation state of 'S' is, (+6)
(b) [tex]Ca(OH)_2[/tex]
Let the oxidation state of 'Ca' be, 'x'
[tex]x+2(-2+1)=0\\\\x=+2[/tex]
Hence, the oxidation state of 'Ca' is, (+2)
(c) [tex]BrOH[/tex]
Let the oxidation state of 'Br' be, 'x'
[tex]x+(-2)+1=0\\\\x=+1[/tex]
Hence, the oxidation state of 'Br' is, (+1)
(d) [tex]ClNO_2[/tex]
Let the oxidation state of 'N' be, 'x'
[tex]-1+x+2(-2)=0\\\\x=+5[/tex]
Hence, the oxidation state of 'N' is, (+5)
(e) [tex]TiCl_4[/tex]
Let the oxidation state of 'Ti' be, 'x'
[tex]x+4(-1)=0\\\\x=+4[/tex]
Hence, the oxidation state of 'Ti' is, (+4)
(f) [tex]NaH[/tex]
Let the oxidation state of 'Na' be, 'x'
[tex]x+(-1)=0\\\\x=+1[/tex]
Hence, the oxidation state of 'Na' is, (+1)
In which of the following bonds, does the Cl atom have the highest electron density? LaTeX: O-Cl O − C l LaTeX: N-Cl N − C l LaTeX: Br-Cl B r − C l LaTeX: H-Cl H − C l LaTeX: S-Cl
hey there!:
Electron density on Cl atom depends on electronegativity difference between Cl and other bonded atom. If the electrnegativity difference is more then Cl has greater electron density, that menas if the bonded atom has less electronegativity then bonded electrons are more attracted by Cl and it has greater electron density.
Among the five atoms which are bonded to Cl atom H has low electronegativity. So in H-Cl two bonded electrons are closer to Cl atom as it has greater electronegativity than H. This results more electron density on Cl atom.
Hence in H-Cl bond Cl atom have the highest electron density
Hope this helps!
The bond in which the chlorine atom has the highest electron density is H − C l.
The polarity of a bond depends on the magnitude of electronegativity difference between the atoms in the bond. The greater the electronegativity difference between the atoms in a bond the more the polarity of the bond.
The magnitude of electron density on each atom in a bond depends on its electronegativity. The more electronegative an atom is, the more it is able to accommodate larger electron density. Looking at the options, hydrogen is far less electronegative than chlorine so a large magnitude of electron density resides on the chlorine atom in HCl.
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Which group 1 element reacts the most vigorously?
Rb
Cs
K
Li
Na
Answer:
The answer is lithium(LI)
In reverse osmosis, water flows out of a salt solution until the osmotic pressure of the solution equals the applied pressure. If a pressure of 70.0 bar is applied to seawater, what will be the final concentration of the seawater at 20 ∘C when reverse osmosis stops?
Answer : The final concentration of the seawater is, 2.909 mole/L
Explanation :
Formula used for osmotic pressure :
[tex]\pi=CRT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure = 70.0 bar = 70 atm
R = solution constant = 0.0821 Latm/moleK
T= temperature of solution = [tex]20^oC=273+20=293K[/tex]
C = final concentration of seawater = ?
Now put all the given values in the above formula, we get the concentration of seawater.
[tex]70atm=C\times 0.0821Latm/moleK\times 293K[/tex]
[tex]C=2.909mole/L[/tex]
Therefore, the final concentration of the seawater is, 2.909 mole/L
To determine the final concentration of seawater when reverse osmosis stops at a pressure of 70.0 bar and 20 °C, additional data such as the initial osmotic pressure of the seawater is required. Without this data, the final concentration cannot be calculated.
In reverse osmosis, water purification occurs by forcing water from a more concentrated solution to a less concentrated one by applying pressure greater than the osmotic pressure. When a pressure of 70.0 bar is applied to seawater at 20 °C, the process will continue until the osmotic pressure of the sea water is equal to the applied pressure. Since the question asks for the final concentration at which the reverse osmosis stops, we would need information about the initial osmotic pressure of seawater to calculate the final concentration. Typically, however, this value can vary, and since the necessary data to perform the calculation is not provided, we cannot accurately provide the final concentration of the seawater.
Reverse osmosis systems, such as those used in desalination plants, continuously introduce seawater under pressure and collect pure water, hence the process carries on indefinitely and the actual concentration in the plants would be constantly changing based on the amount of seawater processed and pure water extracted.
If 25 kg of ice at 0C is combined with 4 kg of steam at 100C, what will be the final equilibrium temperature (in C) of the system? Latent heat of fusion of ice = 3.34 x 105 J.kg-1 Latent heat of steam = 2.23 x 106 J.kg-1 Specific heat of water = 4180 J.kg-1.K-1
Try the suggested option; answer is marked with red colour (18.4953 °C).
All the details are in the attached picture.
The equilibrium temperature of the system is required.
The equilibrium temperature of the mixture is [tex]18.48^{\circ}\text{C}[/tex].
[tex]m_i[/tex] = Mass of ice = 25 kg
[tex]m_s[/tex] = Mass of steam = 4 kg
[tex]T_i[/tex] = Temperature of ice = [tex]0^{\circ}\text{C}[/tex]
[tex]T_s[/tex] = Temperature of steam = [tex]100^{\circ}\text{C}[/tex]
[tex]L_f[/tex] = Latent heat of fusion = [tex]3.34\times 10^5\ \text{J/kg}[/tex]
[tex]L_v[/tex] = Latent heat of vaporization = [tex]2.23\times 10^6\ \text{J/kg}[/tex]
[tex]c_w[/tex] = Specific heat of water = [tex]4180\ \text{J/kg}^{\circ}\text{C}[/tex]
The heat balance of the system will be
[tex]m_iL_f+m_ic_w(T-T_i)=m_sL_v+m_sc_w(T_s-T)\\\Rightarrow m_iL_f+m_ic_wT-m_ic_wT_i=m_sL_v+m_sc_wT_s-m_sc_wT\\\Rightarrow T=\dfrac{m_ic_wT_i+m_sL_v+m_sc_wT_s-m_iL_f}{m_ic_w+m_sc_w}\\\Rightarrow T=\dfrac{25\times 4180\times 0+4\times 2.23\times 10^6+4\times 4180\times 100-25\times 3.34\times 10^5}{25\times 4180+4\times 4180}\\\Rightarrow T=18.49^{\circ}\text{C}[/tex]
The equilibrium temperature of the mixture is [tex]18.49^{\circ}\text{C}[/tex].
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At the beginning of an experiment, the concentration of nitrogen dioxide in a reaction vessel was 0.1103 mol/L. After 65.0 s, the concentration decreased to 0.1076 mol/L. What is the average rate of decomposition of NO2 during this time interval, in mol/(L ∙ s)?
Answer:
-4.15*[tex]10^{-5}mol/(L.s)[/tex]
Explanation:
Average rate = [tex]\frac{final concentration -initial concentration}{change in time} = \frac{0.1076mol/L-0.1103mol/L}{65s}=-4.15.10^{-5}mol/(L.s)[/tex]
The rate is negative because it is a decomposition and our focus is the reactant which is depleting.
The standard reduction potentials of lithium metal and chlorine gas are as follows: Reaction Reduction potential (V) Li+(aq)+e−→Li(s) −3.04 Cl2(g)+2e−→2Cl−(aq) +1.36 In a galvanic cell, the two half-reactions combine to 2Li(s)+Cl2(g)→2Li+(aq)+2Cl−(aq) Calculate the cell potential of this reaction under standard reaction conditions.
Under standard conditions :
E(cell) = E(cathode) - E(anode)
Note : cathode has the larger numeric value and anode has the smaller. Therefore
E(cell) = +1.36V - ( -3.04V)
= 1.36 + 3.04
= +4.40V
In this type of inhibition, the inhibitor can bind to the enzyme regardless of whether the substrate has already bound to the enzyme or not. b) noncompetitive c) uncompetitive d) All of the above a) competitive e) None of the above
Answer:
The correct option is : b) noncompetitive
Explanation:
There are three main types of inhibition:
1. Competitive: In this inhibition, the inhibitor molecule competes with the substrate to bind on the active site of the enzyme.
2. Uncompetitive: In this inhibition, the inhibitor molecule binds to the enzyme- substrate activated complex and thus, does not compete with the substrate to bind on the active site of the enzyme.
3. Non-competitive: In this inhibition, the inhibitor molecule can bind to both the enzyme molecule or to the enzyme-substrate activated complex.
Therefore, In non- competitive inhibition, the inhibitor molecule binds to the enzyme regardless of whether the substrate molecule is bound to the enzyme active site or not.
A simple way to estimate whether a molecule is a strong electrolyte, a weak electrolyte, or a nonelectrolyte is to examine an equation that accurately describes its behavior in water. When examining the chemical equations, start by looking at the direction and type of reaction arrow. A chemical equation reported in the literature may include a forward reaction arrow (→), equilibrium reaction arrow (⇌), or reverse reaction arrow (←). Next, determine whether ions are likely to be present and, if present, in what quantities they should be present.
Chemical equations are symbolic representations of chemical reactions where the reactant entity is given on the left side and the product entity is on the right side. The coefficient next to the symbol and entity formula is the absolute value of the stoichiometric number.
Further ExplanationThe arrow symbol that is usually used is used to distinguish between different types of reactions. To show the type of reaction.
The chemical reaction formula shows the process of how one thing becomes another. Most often, this is written in the format:
Reactants → Products
The right arrow is the most common arrow in a chemical reaction formula. The direction shows the direction of the reaction.
Double arrows indicate reversible reactions. Reactants become products and products can become reactants again using the same process.
Two arrows with a single thorn pointing in the opposite direction show a reversible reaction if the reaction is in equilibrium.
This arrow is used to show the equilibrium reaction where again the arrow points to the stronger side of the reaction.
The reaction above shows a product that is stronger than the reactants. The lower reaction shows the reactants are preferred over the product.
Single Double Arrows are used to indicate resonance between two molecules.
Typically, the reactants will be the resonant isomers of the product.
Curved arrows with single spines on arrows indicate the path of electrons in the reaction. Electrons move from tail to head.
Curved arrows are usually displayed on individual atoms in a skeletal structure to indicate where the electron will be moved from within the product molecule.
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Grade: College
Subject: Chemistry
keywords: Chemical equations
Final answer:
When examining a chemical equation that describes a molecule's behavior in water, you can determine whether it is a strong electrolyte, weak electrolyte, or nonelectrolyte. Strong electrolytes dissociate completely into ions when dissolved, weak electrolytes only partially ionize, and nonelectrolytes do not yield ions.
Explanation:
A simple way to estimate whether a molecule is a strong electrolyte, a weak electrolyte, or a nonelectrolyte is to examine the chemical equation that accurately describes its behavior in water. The direction and type of reaction arrow in the equation can provide valuable information. A chemical equation reported in the literature may include a forward reaction arrow (→), equilibrium reaction arrow (⇌), or reverse reaction arrow (←). By determining whether ions are likely to be present and in what quantities, you can classify the molecule.
Strong electrolytes are substances that dissociate completely into ions when dissolved, resulting in a larger number of dissolved particles and high conductivity. Examples include ionic compounds like sodium chloride (NaCl).
Weak electrolytes only partially ionize when dissolved, resulting in a smaller fraction of ions and lower conductivity. Examples include weak acids like acetic acid (CH3COOH).
Nonelectrolytes do not yield ions when dissolved in water and do not conduct electricity. Examples include molecular compounds like glucose (C6H12O6).