Explanation:
The integrated rate law for the zeroth order reaction is:
[tex][A]=-kt+[A]_0[/tex]
The integrated rate law for the first order reaction is:
[tex][A]=[A]_0e^{-kt}[/tex]
The integrated rate law for the second order reaction is:
[tex]\frac{1}{[A]}=kt+\frac{1}{[A]_0}[/tex]
Where,
[tex][A][/tex] is the active concentration of A at time t
[tex][A]_0[/tex] is the active initial concentration of A
t is the time
k is the rate constant
Answer:
- 0th: [tex]C_A=C_{A0}-kt[/tex]
- 1st: [tex]C_A=C_{A0}exp(-kt)[/tex]
- 2nd: [tex]\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]
Explanation:
Hello,
For the ideal reaction A→B:
- Zeroth order rate law: in this case, we assume that the concentration of the reactants is not included in the rate law, therefore the integrated rate law is:
[tex]\frac{dC_A}{dt}=-k\\ \int\limits^{C_A}_{C_{A0}} {} \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\C_A-C_{A0}=-kt\\C_A=C_{A0}-kt[/tex]
- First order rate law: in this case, we assume that the concentration of the reactant is included lineally in the rate law, therefore the integrated rate law is:
[tex]\frac{dC_A}{dt}=-kC_A\\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A} } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\ln(\frac{C_{A}}{C_{A0}} )=-kt\\C_A=C_{A0}exp(-kt)[/tex]
- Second order rate law: in this case, we assume that the concentration of the reactant is squared in the rate law, therefore the integrated rate law is
[tex]\frac{dC_A}{dt}=-kC_A^{2} \\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A^{2} } } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\-\frac{1}{C_A}+\frac{1}{C_{A0}}=-kt\\\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]
Best regards.
Trinitrotoluene (C-H5N306, 227.1 g/mol) is easily detonated. How many grams of carbon are in 57.6 grams of TNT? Avogadro's Number: 1 mole = 6.02 x 1023 species A. 403 g B. 57.6 g C. 21.3 g D. 1.78 g E. None of the above
Answer: The correct answer is Option C.
Explanation:
The chemical formula for trinitrotoluene is [tex]C_7H_5N_3O_6[/tex]
In 1 mole of TNT, 7 moles of carbon atom, 5 moles of hydrogen atom, 3 moles of nitrogen atom and 6 moles of oxygen atom are present.
We know that:
Mass of trinitrotoluene = 227.1 g/mol
Mass of carbon = 12.01 g/mol
We are given:
Mass of TNT = 57.6 grams
To calculate the mass of carbon in given amount of TNT, we apply unitary method:
In 227.1 grams of TNT, amount of carbon present is = [tex](7\times 12.01)=84.07g[/tex]
So, in 57.6 grams of TNT, the amount of carbon present is = [tex]\frac{84.07g}{227.1g}\times 57.6g=21.3g[/tex]
Hence, the correct answer is Option C.
A buffer, consisting ofH2PO4-
andHPO42-, helps control the pH of
physiologicalfluids. Many carbonated soft drinks also use this
buffer system.What is the pH of a soft drink in which the major
bufferingredients are 6.70 g ofNaH2PO4 and
6.50g of Na2HPO4 per 355 mL
ofsolution?
Answer:
The pH of the drink is 7.12
Explanation:
First, we calculate the concentration of NaH₂PO₄ and Na₂HPO₄, using their molecular weight and the volume in L (355 mL= 0.355 L):
[NaH₂PO₄] = [tex]\frac{6.70g}{0.355L*120g/mol}= 0.1573 M[/tex]
[Na₂HPO₄] = [tex]\frac{6.50g}{0.355L*142g/mol} = 0.1289 M[/tex]
Now we calculate the pH of the solution, keeping in mind the equilibrium:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺From literature, we know that the pka for the previous equilibrium is 7.21
The equation that gives us the pH of a buffer solution is the Henderson–Hasselbalch equation:
pH = pka + [tex]log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}[/tex]
Replacing in the equation the data we know gives us:
[tex]pH=7.21+log\frac{0.1289M}{0.1573M} \\pH=7.12[/tex]
Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction: 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) Imagine that you have a 5 L gas tank and a 3.5 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 127 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Answer:
72.6 atm should be the pressure at which acetylene tank.
Explanation:
[tex]2C_2H_2 (g)+5O_2 (g)\rightarrow 4CO_2 (g)+2H_2O[/tex]
Let the temperature of the both tanks be same as T.
Volume of the tank in which oxygen is filled = [tex]V_1=5L[/tex]
Pressure of the oxygen in tank =[tex]P_1[/tex]= 127 atm
According to reaction 5 moles of oxygen reacts with 2 moles of acetylene.
[tex]n_1= 5 moles[/tex]
[tex]P_1V_1=n_1RT[/tex]
[tex]T=\frac{P_1V_1}{n_1}[/tex]..[1]
[tex]n_2=2 mol[/tex]
Volume of the tank in which acetylene is filled = [tex]V_2=3.5L[/tex]
Pressure of the acetylene in tank =[tex]P_2[/tex]= ?
[tex]T=\frac{P_2V_2}{n_2}[/tex]..[2]
[1] = [2]
[tex]\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}[/tex]
[tex]P_2=\frac{P_1\times V_1\times n_2}{n_1\times V_2}[/tex]
[tex]=\frac{127 atm\times 5 L \times 2 mol}{5 mol\times 3.5 L}[/tex]
[tex]P_2=72.6 atm[/tex]
72.6 atm should be the pressure at which acetylene tank.
Which one of the following statements is not correct?
Select one:
a. During allotropic transformation, the number of atoms in the material changes.
b. During allotropic transformation the volume of the material changes.
c. Isotopes of the same element have the same number of protons.
d. In an element, the size of its anion is larger than its atomic size
Answer:
option a, During allotropic transformation, the number of atoms in the material changes.
Explanation:
Allotrops are different crystalline forms of the same element. So, two allotrops of an element have same number of atoms.
for example diamond, graphite, graphene, fullerene are allotropic forms of carbon.
Allpotops differ in physical and chemical properties. So, volume changes during alloptropic transformation. So, statement b is correct.
Allotropic transformation is the transformation of one allotropic form to other therefore, number of atom does not change during allotropic transformation.
Chemical species having same atomic number or same no. of protons are called isotopes. so statement c is correct.
When an electron is added to the neutral element, electronic repulsion increases which lead to the increase in atomic size, So, statement d is also correct.
So among given, option a is incorrect
Evaporation of Cane Sugar Solutions. An evaporator is used to concentrate cane sugar solutions. A feed of 10 000 kg/d of a solution containing 38 wt % sugar is evaporated, producing a 74 wt % solution. Calculate the weight of solution produced and amount of water removed.
Answer:
Weight of solution produced = 5135 kg
Amount of water removed = 4865 kg
Explanation:
For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg
mi = 0.38x10000 = 3800 kg
The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)
mo = 0.74xmt
0.74xmt = 3800
mt = 3800/0.74
mt = 5135 kg
This is the mass of solution produced.
The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :
wi = 10000 - 3800 = 6200 kg
wo = 5135 - 3800 = 1335 kg
wr = wi - wo
wr = 6200 - 1335
wr = 4865 kg
From the data provided and the concept of balance of mass,, the mass of solution produced is 5135 kg and the water removed is 4865 kg
What is evaporation?
Evaporation is the process by which molecules of a liquid turn to gas.
To calculate the weight of solution produced and amount of water removed:
Using the concept of balance of mass, the incoming mass of sugar must be equal to the outgoing mass.
Incoming mass (Mi) = 38% of 10000 kg
Mi = 3800 kg
Therefore, the outgoing mass (Mo)= 3800 kg
Mo= 74% of the total mass, Mt
Mo = 0.74 x Mt
Mt = 3800/0.74
Mt = 5135 kg
Thus, the mass of solution produced is 5135 kg
The amount of water removed (Wr) = water incoming (Wi) - water outgoing (wo).
Wi = 10000 - 3800 = 6200 kg
Wo = 5135 - 3800 = 1335 kg
Wr = Wi - Wo
Wr = 6200 - 1335
Wr = 4865 kg
Therefore, the water removed is 4865 kg
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Convert the condensed structures to line angle formulas: 1. CH3CH2CH(CH3)CH2CH3 8. CH:COCH 9. CH3CH2OCH3 2. CH3CH2CH(CH2CH3)CH(CH3)CH2CHO CH)CH:CH-CH: 10. CH CH2CH=C(CH:CH)(C(CH))CH 3. CH3CH2CH(CH3)CH(CH3)CH2CH3 11. CH:CH-CH(CH2CH)CH OH 4. CH3C(CH3)2CH2CH2CH(CH3)CH2CH3 12. CHCHOCH2CHO 5. CH(CH3)2CH(CI)CH2CH3 13. HOOCCH2NHCH(CH3)COCH; 6. CH3CH2CHOCH(CH2CH3)CH2CH3 14. HOOCCH OCH COOH 7. HOCH:C(CH3)2CONH2
Answer:
Attached in the photo.
Explanation:
Hello,
The answers in the attached photo. Just three things:
- In the second point a parenthesis is missing to properly understand the molecule (after the oxygen), nevertheless, I assumed it was an ether.
- In the sixth point, there's a missing hydrogen for it to be an ether as well.
- In the tenth point the second parenthesis is not clear, it seems there's a missing subscript, nevertheless I draw it assuming complete octates.
Best regards.
Describe the molecular structure of water and explain why the water molecule is polar
Explanation:
Hybridization of O in [tex]H_2O = sp^3[/tex]
So, water molecule has four hybrid orbitals.
Two hybrid orbitals form 2 sigma bond with two H atoms.
Remaining two hybrid orbitals are occupied by two lone pairs.
Because of lone pair-lone pair repulsion, shape of [tex]H_2O[/tex] becomes bent.
Water molecule is polar because of difference in eletronegativities of O and H.
O is more electronegative as comapared to hydrogen. So bonding electrons get attracted towards O atom which results in the development of partial negative charge on O atom and partial positive charge on H atoms.
Because of development of partial negative and partial positive charge, water molecule becomes polar.
Water's molecular structure is a bent shape with two hydrogen atoms bonded to one oxygen atom. The electronegativity difference between oxygen and hydrogen results in a polar molecule, leading to properties like high boiling point and solubility.
Explanation:The molecular structure of water, referred to as H2O, comprises two hydrogen atoms bonded to one oxygen atom. Each hydrogen atom forms a single covalent bond with the oxygen atom, creating a bent structure. Yet, oxygen is more electronegative than hydrogen, meaning it pulls shared electrons closer to itself. This results in oxygen having a partial negative charge, and hydrogen having a partial positive charge, making water a polar molecule.
Water's polarity contributes to its unique properties, such as high boiling point and ability to dissolve many substances. Water molecules can form hydrogen bonds - attractions between the positively charged hydrogen of one molecule, and the negatively charged oxygen of another - due to their polarity, which makes the water molecule extremely cohesive and leads to a higher than expected boiling point.
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Which ionic compound would you expect to have the highest melting point? a. Al-N b. Fe-O C. W-C
Answer:
b. Fe-O
Explanation:
A way to predict melting points for ionic compounds is with electronegativity.
Electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.
The most electronegativity difference (E.D.), the highest melting point, thus:
E.D. Al-N = 3,0 - 1,6 = 1,4
E.D. Fe-O = 3,5 - 1,8 = 1,7
E.D. W-C = 2,5 - 1,7 = 0,8
The most electronegativity difference is from Fe-O, thus, this ionic compound will have the highest melting point.
I hope it helps!
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Using the Supplemental Data, determine how much heat (in kJ) is produced by the decomposition of 1.05 mol of H2O2 under standard conditions.
2 H2O2(l) → 2 H2O(g) + O2(g)
Answer:
The heat released is 56.7 kJ.
Explanation:
To solve this problem, first we need to find out the standard enthalpy of reaction, that is, the energy released at constant pressure in standard conditions (P=1bar, T=298.15K). We can find it using the expression:
ΔH°r = Σn(p).ΔH°f(p) - Σn(r).ΔH°f(r)
where,
n refers to the number of moles of reactants and products in the balanced equation
ΔH°f refers to standard enthalpies of formation (which can be found in tables).
Given the equation:
2 H₂O₂(l) → 2 H₂O(g) + O₂(g)
We can replace with the proper data in the equation:
ΔH°r = Σn(p).ΔH°f(p) - Σn(r).ΔH°f(r)
ΔH°r= [2 mol . ΔH°f H₂O(g) + 1 mol . ΔH°f O₂(g)] - [2 mol . ΔH°f H₂O₂(l)]
ΔH°r= [2 mol . (-241.8 kJ/mol) + 1 mol . 0 kJ/mol] - [2 mol . (-187.8 kJ/mol)]
ΔH°r = -108.0 kJ
Since enthalpy is an extensive property, it depends on the amount of reagents. In this case, 108.0 kJ of heat are released every 2 moles of H₂O₂(l) decomposed. Then, for 1.05 mol of H₂O₂(l):
[tex]1.05 mol.\frac{-108.0kJ}{2mol} =-56.7kJ[/tex]
By convention, the negative sign means that heat is released.
You make 1 Liter of an aqueous solution containing 9.20 ml of 57.8 mM acetic acid and 56.2 mg of sodium acetate (MW = 82.0 g/mole). a. How many moles of acetic acid did you add? b. How many moles of sodium acetate did you add? c. What is the appropriate formula for calculating the pH of the above solution? d. What is the pH of the above solution? Is it acidic or basic? Slightly or strongly so? (Use the same scale as on HW 1)
Answer:
a) 5,3176x10⁻⁴ moles
b) 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch.
d) pH = 4,86. Acidic solution but slighty
Explanation:
a) moles of acetic acid:
9,20x10⁻³L × 57,8x10⁻³M = 5,3176x10⁻⁴ moles
b) moles of sodium acetate:
56,2x10⁻³g ÷ 82,0 g/mole = 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch:
pH= pka + log₁₀ [tex]\frac{[A^-]}{[HA]}[/tex]
d) pH= 4,75 + log₁₀ [tex]\frac{[6,85x10_{-4}]}{[5,3176x10_{-4}]}[/tex]
pH = 4,86
3 < pH < 7→ Acidic solution but slighty
I hope it helps!
According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. A 150-lbm swimmer is floating in nearby pool; 95% of her body's volume is in the water while 5% of her body's volume is above water. Determine the density of the swimmer's body The density of water is 0.036 lbm/in5. Does your answer make sense?? Why why not?
Answer:
The density of the swimmer is 0.0342 lbm/in3.
This value makes sense as the density of the body is very similar to the water.
Explanation:
If the swimmers is floating, the weight of the swimmer must be equal to the upward buoyant force.
We can express the weight force as the product of density and volume of the swimmer.
Then
[tex]\rho_{swimmer}*V_{swimmer}=\rho_{water}*V_{water_displaced}\\\\
\rho_{swimmer}*V_{swimmer}=\rho_{water}*0.95**V_{swimmer}\\\\
\rho_{swimmer}=0.95*\rho_{water}\\\\
\rho_{swimmer}=0.95*0.036 lbm/in3\\\\
\rho_{swimmer}=0.0342lbm/in3[/tex]
It makes sense as the density of the body is very similar to the water.
A solution is to be prepared with a sodium ion concentrationof
0.148 mol/L. What mass of sodium sulfate (g) is needed toprepare
2.29 liters of such a solution?
Answer : The mass of sodium sulfate needed is 5.7085 grams.
Explanation : Given,
Concentration of sodium ion = 0.148 mol/L
Volume of solution = 2.29 L
Molar mass of sodium sulfate = 142 g/mole
First we have to determine the moles of sodium ion.
[tex]\text{Concentration of sodium ion}=\frac{\text{Moles of sodium ion}}{\text{Volume of solution}}[/tex]
[tex]0.184mol/L=\frac{\text{Moles of sodium ion}}{2.29L}[/tex]
[tex]\text{Moles of sodium ion}=0.08035mole[/tex]
Now we have to calculate the moles of sodium sulfate.
The balanced chemical reaction will be,
[tex]Na_2SO_4\rightarrow 2Na^++SO_4^{2-}[/tex]
As, 2 moles of sodium ion produced from 1 moles of [tex]Na_2SO_4[/tex]
So, 0.08035 moles of sodium ion produced from [tex]\frac{0.08035}{2}=0.040175[/tex] moles of [tex]Na_2SO_4[/tex]
Now we have to calculate the mass of sodium sulfate.
[tex]\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4\times \text{Molar mass of }Na_2SO_4[/tex]
[tex]\text{Mass of }Na_2SO_4=0.040175mole\times 142g/mole=5.7085g[/tex]
Therefore, the mass of sodium sulfate needed is 5.7085 grams.
A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a) 20 b) 9.25 c) 25 d) 6.75
Answer: The correct answer is Option d.
Explanation:
We are given:
Mass percentage of [tex]CH_4[/tex] = 20 %
So, mole fraction of [tex]CH_4[/tex] = 0.2
Mass percentage of [tex]C_2H_4[/tex] = 30 %
So, mole fraction of [tex]C_2H_4[/tex] = 0.3
Mass percentage of [tex]C_2H_2[/tex] = 35 %
So, mole fraction of [tex]C_2H_2[/tex] = 0.35
Mass percentage of [tex]C_2H_2O[/tex] = 15 %
So, mole fraction of [tex]C_2H_2O[/tex] = 0.15
We know that:
Molar mass of [tex]CH_4[/tex] = 16 g/mol
Molar mass of [tex]C_2H_4[/tex] = 28 g/mol
Molar mass of [tex]C_2H_2[/tex] = 26 g/mol
Molar mass of [tex]C_2H_2O[/tex] = 48 g/mol
To calculate the average molecular mass of the mixture, we use the equation:
[tex]\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}[/tex]
where,
[tex]\chi_i[/tex] = mole fractions of i-th species
[tex]m_i[/tex] = molar masses of i-th species
[tex]n_i[/tex] = number of observations
Putting values in above equation:
[tex]\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}[/tex]
[tex]\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75[/tex]
Hence, the correct answer is Option d.
The average molecular weight of the gaseous mixture is approximately 27 g/mol, which is not one of the provided options.
The average molecular weight (MW) of a mixture can be calculated as the sum of the weight percentages of each component multiplied by its respective molecular weight divided by 100. For the given mixture, the molecular weights of the components are:
CH4 = 16 g/mol
C2H4 = 28 g/mol
C2H2 = 26 g/mol
C2H2O = 42 g/mol.
We multiply the weight percentages with the molecular weights of each component and then sum them up:
(20% x 16 g/mol) + (30% x 28 g/mol) + (35% x 26 g/mol) + (15% x 42 g/mol)=(0.20 x 16) + (0.30 x 28) + (0.35 x 26) + (0.15 x 42)=3.2 + 8.4 + 9.1 + 6.3=27 g/molHence, the average molecular weight of the mixture is 27 g/mol. Thus, none of the options provided (a-d) is correct.
(a) Consider a carbon atom in its ground state. Would such an atom offer a satisfactory model for the carbon of methane? If not, why not? (Hint: Consider whether a ground state carbon atom could be tetravalent, and consider the bond angles that would result if it were to combine with hydrogen atoms.)
(b) Consider a carbon atom in the excited state. Would such an atom offer a satisfactory model for the carbon of methane? If not, why not?
Answer:
(a) No. Ground-state carbon has only 2 half-filled orbitals that could be used for bonding.
(b) No. The bond angles would be incorrect as the p-orbitals are all perpendicular to each other (90°).
Explanation:
See attachment for the ground-state and excited-state electron orbital diagrams of carbon.
A methane molecule has all four CH bonds the same length and at 109.5° from each other. Hybridization of the s and p orbitals to sp³ orbitals is necessary.
have an infinite number of significant figures
Exact numbers have an infinite number of significant figures because the number of significant figures in a value indicates the level of uncertainty associated with that value, and exact numbers have no associated uncertainty.
What are exact numbers ?
Exact numbers are those whose values are known with absolute certainty, devoid of any measurement uncertainty or approximation. They are typically derived from precise definitions, counting procedures, or other mathematical operations that guarantee their accuracy.
Due to their certainty, exact numbers are considered to possess an infinite number of significant figures. This is because the concept of significant figures reflects the level of uncertainty in a value, and exact numbers, being devoid of uncertainty, transcend this limitation.
Complete question:
Why do exact numbers have an infinite number of significant figures?
A chemist measures the amount of hydrogen gas produced during an experiment. She finds that 264. g of hydrogen gas is produced. Calculate the number of moles of hydrogen gas produced. Round your answer to 3 significant digits. W mol x 6 ?
Answer:
The answer is 130.953 g of hydrogen gas.
Explanation:
Hydrogen gas is formed by two atoms of hydrogen (H), so its molecular formula is H₂. We can calculate is molecular weight as the product of the molar mass of H (1.008 g/mol):
Molecular weight H₂= molar mass of H x 2= 1.008 g/mol x 2= 2.01568 g
Finally, we obtain the number of mol of H₂ there is in the produced gas mass (264 g) by using the molecular weight as follows:
mass= 264 g x 1 mol H₂/2.01568 g= 130.9731703 g
The final mass rounded to 3 significant digits is 130.973 g
To find the number of moles of hydrogen gas produced from 264 grams, divide the mass by the molar mass of hydrogen (2.02 g/mol), resulting in approximately 130.7 moles of hydrogen gas to three significant digits.
Explanation:To calculate the number of moles of hydrogen gas (H₂) produced from 264 grams of hydrogen gas, you would use the molar mass of H₂ which is approximately 2.02 g/mol (1 mole of H₂ = 2.02 grams). Using the formula:
number of moles = mass of substance (g) / molar mass (g/mol)
We find the number of moles of hydrogen gas to be:
number of moles = 264 g / 2.02 g/mol
After performing the division, this gives us approximately 130.7 moles of H₂.
This result is to three significant digits, aligned with the precision provided by the initial mass of the hydrogen gas.
A temperature difference of 15°C is impressed across a brick wall of a house which is 15 cm in thickness. The thermal conductivity of the brick is 1.0 W/m °C. The face of the wall is 6 meters high and 12 meters long. Compute both the heat flux and the heat transfer rate through the wall. Why aren't these numerical values the same?
Answer:
Q=7200 W
q=7200/72=100 [tex]W/m^2[/tex]
Explanation:
Given that
ΔT=15° C
Thickness ,t=15 cm
Thermal conductivity ,K=1 W/m.°C
Height,h=6 m
Length ,L=12 m
As we know that heat conduction through wall given as
[tex]Q=\dfrac{KA}{t}\Delta T[/tex]
Now by putting the values
A= 6 x 12 =72 [tex]m^2[/tex]
[tex]Q=\dfrac{KA}{t}\Delta T[/tex]
[tex]Q=\dfrac{1\times 72}{0.15}\times 15\ W[/tex]
Q=7200 W
Q is the total heat transfer.
Heat flux q
q=Q/A [tex]W/m^2[/tex]
q=7200/72=100 [tex]W/m^2[/tex]
q is the heat flux.
As w know that heat flux(q) is the heat transfer rate from per unit area and on the other hand heat transfer(Q) is the total heat transfer from the surface.
Heat flux q=Q/A
That is why these both are different.
Which of the following gives the net ionic reaction for the reaction used in this experiment?
H+(aq) + OH-(aq) → H2O(l)
no net reaction
Ba2+(aq) + SO42-(aq) → BaSO4(s)
Ba2+(aq) + 2 OH-(aq) + 2 H+(aq) + SO42-(aq) → BaSO4(s) + 2 H2O(l)
Answer: Option (d) is the correct answer.
Explanation:
The given reaction will be as follows.
[tex]Ba(OH)_{2}(aq) + H_{2}SO_{4}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]
In the ionic form, the equation will be as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) \rightarrow BaSO_{4}(s)[/tex] ........ (1)
[tex]H^{+}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l)[/tex] ............ (2)
Hence, for the net ionic equation we need to add both equation (1) and (2). Therefore, the net ionic equation will be as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + H^{+}(aq) + OH^{-}(aq) \rightarrow BaSO_{4}(s) + H_{2}O(l)[/tex]
Now, balancing the atoms on both the sides we get the net ionic equation as follows.
[tex]Ba^{2+}(aq) + SO^{2-}_{4}(aq) + 2H^{+}(aq) + 2OH^{-}(aq) \rightarrow BaSO_{4}(s) + 2H_{2}O(l)[/tex]
The correct net ionic reaction for the experiment is H+(aq) + OH-(aq) → H2O(l), an example of an acid-base neutralization reaction. The other options included the formation of a precipitate, which is not part of the net ionic reaction.
Explanation:Based on the provided options, the correct net ionic reaction for the experiment seems to be H+(aq) + OH-(aq) → H2O(l). This reaction is a classic example of an acid-base neutralization reaction, where an acid (H+) and a base (OH-) react to form water. The other two reactions involve the formation of a precipitate (BaSO4), but the full reaction is simplified to leave out the precipitate ions on either side. This does not occur in the first reaction. Hence, the first reaction is the correct net ionic reaction for this experiment.
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Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, this type of reaction is used to produce products such as margarine. A typical hydrogenation reaction is C10H20() + H2(g) → C10H22(5) Decene Decane How much decane can be produced in a reaction of excess decene with 2.45 g hydrogen? Give your answer in scientific notation. O *10 g decane
Answer: The mass of decane produced is [tex]1.743\times 10^2g[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Mass of hydrogen gas = 2.45 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1:, we get:
[tex]\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol[/tex]
The chemical equation for the hydrogenation of decene follows:
[tex]C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)[/tex]
As, decene is present in excess. So, it is considered as an excess reagent.
Thus, hydrogen gas is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of hydrogen gas produces 1 mole of decane.
So, 1.225 moles of hydrogen gas will produce = [tex]\frac{1}{1}\times 1.225=1.225mol[/tex] of decane
Now, calculating the mass of decane by using equation 1, we get:
Moles of decane = 1.225 mol
Molar mass of decane = 142.30 g/mol
Putting values in equation 1, we get:
[tex]1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g[/tex]
Hence, the mass of decane produced is [tex]1.743\times 10^2g[/tex]
How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn3O4 at standard state condaition?
The heat released (ΔH) when one mole of manganese is burned to form Mn3O4 can be derived from a thermochemical equation and molar enthalpies. For specific values, consult a standard enthalpy of formation table.
Explanation:The calculation for the heat released when one mole of manganese is burned to form Mn3O4 at standard state condition involves understanding of chemistry concepts such as molar enthalpies and thermochemical equations. Unfortunately, without specifying the thermochemical equation for the formation of Mn3O4 from manganese or the molar enthalpy of this specific reaction, an exact value cannot be given.
However, as a general guide, the heat released (also known as enthalpy change, ΔH) can be found from the formula ΔH = -q, where q represents the heat absorbed. A negative value indicates heat is being released. In thermochemical equations, the ΔH value is often given per mole of a substance involved in the reaction, so you would usually directly obtain the heat released when one mole of a substance is involved from the molar enthalpy.
For specific values, refer to a standard enthalpy of formation table, a resource often found in chemistry textbooks or scientific literature, to find the molar enthalpy for the formation of Mn3O4 from manganese.
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Answer the following and round to the correct number of significant figures (don't forget units).
a) 0.02123 m + 1.12 m + 0.00123 m =
b) 2.3 cm - 1.23 cm + 120 cm =
c) 25,430 km - 3,500 km + 200 km =
d) (1.21 x 105 ) x (2.6 x 103 ) =
e) 7.13 mm x 9.1 mm =
f) 3.0 cm x 8.222 cm =
g) 4.1 g ÷ 0.121 cm =
h) 0.413 ÷ (9.212 x 103 ) =
i) (12.1 cm - 4.15 cm) / 35.64 g =
j) (11.00 m - 3.356 m) x 45.1 kg /35.64 s =
k) 73.0 x 1.340 x (25.31 – 1.6) =
l) (418.7 x 31.8) / (19.27 – 18.98) =
Answer:
a) 0.02123 m + 1.12 m + 0.00123 m = 1.14246m it can be round off as 1.1425m
b) 2.3 cm - 1.23 cm + 120 cm = 121.07 cm it can be round off as 121.1 cm
c) 25,430 km - 3,500 km + 200 km = 22130 km
d) (1.21 x 105 ) x (2.6 x 103 ) = 34023.99 it can be round off as 34024
e) 7.13 mm x 9.1 mm = 64.883 mm it can be round off as 65 mm
f) 3.0 cm x 8.222 cm = 24.666 cm it can be round off as 25 cm
g) 4.1 g ÷ 0.121 cm = 33.8842 g/cm it can round off as 34 g/cm
h) 0.413 ÷ (9.212 x 103 ) = 0.00043527
i) (12.1 cm - 4.15 cm) / 35.64 g = 0.223063973 cm/g
j) (11.00 m - 3.356 m) x 45.1 kg /35.64 s = 9.672962963 m kg/s
k) 73.0 x 1.340 x (25.31 – 1.6) = 2319.3122
l) (418.7 x 31.8) / (19.27 – 18.98) = 45912.62069 45912.6207
The question requests computations involving addition, subtraction, multiplication, and division while carefully applying the rules of significant figures. The answers are expressed in various physical units such as meters, centimeters, km, mm, g, m/s, and their calculated results are in accordance with significant figures rules.
Explanation:This question involves numerical calculations with significant figures focusing on measurement and precision, a concept prevalent in Physics.
a) 0.02123 m + 1.12 m + 0.00123 m = 1.14 mb) 2.3 cm - 1.23 cm + 120 cm = 121 cmc) 25,430 km - 3,500 km + 200 km = 22,100 kmd) (1.21 x 105 ) x (2.6 x 103 ) = 3.14 x 108e) 7.13 mm x 9.1 mm = 65 mm2f) 3.0 cm x 8.222 cm = 25 cm2g) 4.1 g ÷ 0.121 cm = 33.9 g/cmh) 0.413 ÷ (9.212 x 103 ) = 4.48 x 10-5i) (12.1 cm - 4.15 cm) / 35.64 g = 0.223 g/cmj) (11.00 m - 3.356 m) x 45.1 kg /35.64 s = 9.60 kg.m/sk) 73.0 x 1.340 x (25.31 – 1.6) = 2,300l) (418.7 x 31.8) / (19.27 – 18.98) = 71,900Learn more about Significant Figures here:https://brainly.com/question/37022020
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The standard heat of combustion of liquid ethyl cyclopentane, C7H14(6), was measured to be -4589.6 kJ/mol. What is the standard heat of formation of liquid ethyl cyclopentane?
Answer:
ΔHf C₇H₁₄ = 137.6 kJ/mol
Explanation:
The following equation represents the combustion of liquid cyclopentane:
C₇H₁₄ (l) + 10,5O₂ (g) → 7CO₂ (g) + 7H₂O (g)
The standard heat of combustion for this reaction is ΔHc = -4589.6 kJ/mol.
ΔHc can be calculated by the following equation:
ΔHc = ∑ΔHf products + ∑ΔHf reactants
ΔHc = (7ΔHf CO₂ + 7ΔHf H₂O) - (ΔHf C₇H₁₄ + 10.5ΔHf O₂)
Therefore, we can calculate the standard heat of formation of C₇H₁₄ :
ΔHf C₇H₁₄ = - ΔHc +7ΔHf CO₂+7ΔHf H₂O - 10.5ΔHf O₂
ΔHc = -4589.6 kJ/mol, ΔHf CO₂ = -394.0 kJ/mol, ΔHf H₂O = -242.0 kJ/mol, ΔHf O₂ = 0
ΔHf C₇H₁₄ = 4589.6 + 7x(-394.0) + 7x(-242.0) - 10.5x0
ΔHf C₇H₁₄ = 137.6 kJ/mol
If a chemical reaction consumes reactants at a steady rate of 7.64 x 10^19 molecules per second, how long will it take for the reaction to consume 6.02 x 1017 molecules of reactant?
Answer:
time = Molecules/Rate => 0.0079 segs
Explanation:
Rate = 7.64 * 10^19 molecules/segs
Molecules = 6.02 * 10^17 molecules
time = #?
time = Molecules/Rate => 0.0079 segs
The dimensional analysis calculates the variable from the given data. A chemical reaction with a reaction rate of 7.64 X 10¹⁹ will take 0.0079 seconds to consume 6.02 x 10¹⁷ molecules of reactant.
What is the reaction rate?The reaction rate has been defined by the speed or the time taken for the product to get produced by the reactant undergoing the chemical reaction. The rate of reaction depends on the concentration of the reactants.
Given,
Rate of reaction = 7.64 x 10¹⁹ molecules per second
Molecules = 6.02 × 10¹⁷ molecules
Time is calculated by the dimensional analysis as,
Time = Molecules ÷ Rate
= 6.02 × 10¹⁷ molecules ÷ 7.64 x 10¹⁹ molecules per second
= 0.0079 seconds
Therefore, it will take 0.0079 seconds for 6.02 x 10¹⁷ molecules of reactant to yield the product.
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Balance the equ. representing the reaction between iron
(III)oxide, Fe2O3 and carbon monoxide (CO) to yeild iron (Fe) and
carbondioxide (CO2).
Final answer:
The balanced chemical equation for the reaction between iron (III) oxide (Fe₂O₃) and carbon monoxide (CO) to yield iron (Fe) and carbon dioxide (CO₂) is: Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g).
Explanation:
To balance the equation representing the reaction between iron (III) oxide (Fe₂O₃) and carbon monoxide (CO) to yield iron (Fe) and carbon dioxide (CO₂), we follow the basic principle of conservation of mass, which states that atoms must be conserved in a chemical reaction. The balanced chemical equation for this reaction is:
Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g)
This equation indicates that one mole of iron (III) oxide reacts with three moles of carbon monoxide to produce two moles of iron and three moles of carbon dioxide. Note that this reaction is a combination of reduction of iron oxide and oxidation of carbon monoxide. Such reactions where both oxidation and reduction occur are known as redox reactions.
CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is left in the products?
Answer : The moles of [tex]O_2[/tex] left in the products are 0.16 moles.
Explanation :
First we have to calculate the moles of [tex]CH_4[/tex].
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 1 atm
V = volume of gas = 10 L
T = temperature of gas = [tex]27^oC=273+27=300K[/tex]
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
[tex](1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)[/tex]
[tex]n=0.406mole[/tex]
Now we have to calculate the moles of [tex]O_2[/tex].
The balanced chemical reaction will be:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of [tex]CH_4[/tex] react with 2 moles of [tex]O_2[/tex]
So, 0.406 mole of [tex]CH_4[/tex] react with [tex]2\times 0.406=0.812[/tex] moles of [tex]O_2[/tex]
Now we have to calculate the excess moles of [tex]O_2[/tex].
[tex]O_2[/tex] is 20 % excess. That means,
Excess moles of [tex]O_2[/tex] = [tex]\frac{(100 + 20)}{100}[/tex] × Required moles of [tex]O_2[/tex]
Excess moles of [tex]O_2[/tex] = 1.2 × Required moles of [tex]O_2[/tex]
Excess moles of [tex]O_2[/tex] = 1.2 × 0.812 = 0.97 mole
Now we have to calculate the moles of [tex]O_2[/tex] left in the products.
Moles of [tex]O_2[/tex] left in the products = Excess moles of [tex]O_2[/tex] - Required moles of [tex]O_2[/tex]
Moles of [tex]O_2[/tex] left in the products = 0.97 - 0.812 = 0.16 mole
Therefore, the moles of [tex]O_2[/tex] left in the products are 0.16 moles.
Name the type of bond in organic chemistry that corresponds to a glycoside bond
Answer:
covalent bond
Explanation:
The bond which is most common in the organic molecules is the covalent bond which involves sharing of the electrons between the two atoms.
Glycosidic bond, also known as glycosidic linkage is type of the covalent bond which joins carbohydrate molecule to other group that may not or may be a carbohydrate.
Glycosidic bond is the bond which is formed between hemiacetal or hemiketal group of the saccharide and hydroxyl group of compounds like alcohol.
Lewis base is a(n) a. producer of OH ions. b. proton acceptor. c. electron-pair donor. d. electron-pair acceptor. In the reaction represented by the equation Ag (aq) +2NH3(aq)-[Ag[ a. Bronsted-Lowry acid. b. Lewis acid
Answer:
c. electron-pair donor.
Explanation:
Lewis base -
A Lewis base is an electron rich species , which is available for donation .
Hence , Lewis base is a electron - pair donor .
The indication of a Lewis base , is it has a negative charge or lone pairs of electrons .
Hence , the species with a lone pair or negative charge act as a Lewis base .
for example , OH ⁻ is a Lewis base , due to its negative charge which is available for donation .
H₂O is also a Lewis base , due to its lone pairs of electrons , that are available for donation .
Methyl isocyanate, H3C-N=C=O, is used in the industrial synthesis of a type of pesticide and herbicide known as a carbamate. As a historical note, an industrial accident in Bhopal, India in 1984 resulted in leakage of an unknown quantity of this chemical into the air. An estimated 200,000 persons were exposed to its vapors and over 2000 of these people died.
(a) Draw the Lewis structure for methyl isocyanate.
Explicitly draw all H atoms.
Include all valence lone pairs in your answer.
(b) What is the hybridization of the carbonyl carbon? _________sp²spsp³
What is the hybridization of the nitrogen? _________sp²spsp³
Answer:
(a) The lewis structure for methylisocyanate is in the attached.
(b) The carbonyl carbon have an sp² hybridization
(c) The nitrogen have an sp² hybridization?
Explanation:
(a) The lewis structure for methylisocyanate has the nitrogen with one lone pair and the oxygen with two lone pairs.
(b) The carbonyl carbon form double bond with the oxygen causing to form three hybrid orbitals sp².
The Nitrogen also forms a double bond with the carbon having an sp² hybridization too.
What is the pH of a 0.18 M CH3NH3+Cl– aqueous solution? The pKb ofCH3NH2 is 3.44
Answer:
5.65
Explanation:
Given that:
[tex]pK_{b}\ of\ CH_3NH_2=3.44[/tex]
[tex]K_{b}\ of\ CH_3NH_2=10^{-3.44}=3.6308\times 10^{-4}[/tex]
[tex]K_a\ of\ CH_3NH_3^+Cl^-=\frac {K_w}{K_b}=\frac {10^{-14}}{3.6308\times 10^{-4}}=2.7542\times 10^{-11}[/tex]
Concentration = 0.18 M
Consider the ICE take for the dissociation as:
[tex]CH_3NH_3^+[/tex] ⇄ H⁺ + [tex]CH_3NH_2[/tex]
At t=0 0.18 - -
At t =equilibrium (0.18-x) x x
The expression for dissociation constant of acetic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [CH_3NH_2 \right ]}{[CH_3NH_3^+]}[/tex]
[tex]2.7542\times 10^{-11}=\frac {x^2}{0.18-x}[/tex]
x is very small, so (0.18 - x) ≅ 0.18
Solving for x, we get:
x = 0.2227×10⁻⁵ M
pH = -log[H⁺] = -log(0.2227×10⁻⁵) = 5.65
What volume of phenytoin suspension 30 mg/5 mL is required to be added to a suitable diluent to obtain 150 mL phenytoin suspension 20 mg/5 mL? 100
Explanation:
The given data is as follows.
Concentration of phenytoin suspension stock = 30 mg/5mL
Concentration of phenytoin required = 20 mg/5 mL
Volume of phenytoin required = 150 mL
Volume of phenytoin suspension stock required for dilution will be calculated as follows.
[tex]20 \times \frac{150}{30}[/tex]
= 100 mL
Thus, we can conclude that the volume of phenytoin is 100 mL .
To obtain 150 ml of 20 mg/5 ml of phenytoin suspension from a 30 mg/5 ml concentration, 100 ml of the original suspension should be used and the remaining volume filled with a suitable diluent.
Explanation:
To begin with, let's understand that phenytoin suspension is a medication used to treat seizures. The initial phenytoin concentration provided is 30 mg/5 ml. The required concentration is 20 mg/5 ml in a volume of 150 ml.
First, we need to find out how much total phenytoin we need. Considering 20 mg is required for every 5 ml: (
20 mg/5 ml) x 150 ml = 600 mg. This total amount of phenytoin is present in the stronger concentration of 30 mg/5 ml. To find out the volume of this concentration that we need: 600 mg / (30 mg/5 ml) = 100 ml of phenytoin suspension.
Therefore, we need to take 100 ml of the 30 mg/5 ml phenytoin suspension and add a suitable diluent to achieve a total volume of 150 ml with a concentration of 20 mg/5 ml.
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