Final answer:
Intensive properties are constant regardless of the amount of substance, while extensive properties depend on the amount. Temperature and density are examples of intensive properties, whereas mass and volume are extensive properties.
Explanation:
The difference between an intensive property and an extensive property is fundamental in chemistry and relates to how properties of matter are defined in relation to the amount of substance present. Extensive properties are those that depend on the amount of matter, such as mass and volume. For instance, a larger amount of substance will have a greater mass and volume than a smaller amount. In contrast, intensive properties do not depend on the amount of matter. These properties, such as temperature and density, remain consistent regardless of how much substance you have.
For example, whether you have a gallon or a cup of milk at 20 °C, the temperature is the same because temperature is an intensive property. Similarly, the density of a material is an intensive property because it is defined as the mass per unit volume, a ratio of two extensive properties which effectively 'cancels out' the dependence on quantity.
A solution has a pOH of 7.1 at 10∘C. What is the pH of the solution given that Kw=2.93×10−15 at this temperature? Remember to report your answer with the correct number of significant figures
Answer : The pH of the solution is, 7.4
Explanation : Given,
pOH = 7.1
[tex]K_w=2.93\times 10^{-15}[/tex]
First we have to calculate the value of [tex]pK_w[/tex].
The expression used for the calculation of [tex]pK_w[/tex] is,
[tex]pK_w=-\log [K_w][/tex]
Now put the value of [tex]K_w[/tex] in this expression, we get:
[tex]pK_w=-\log (2.93\times 1-^{-15})[/tex]
[tex]pK_w=15-\log (2.93)[/tex]
[tex]pK_w=14.5[/tex]
Now we have to calculate the pH of the solution.
As we know that,
[tex]pH+pOH=pK_w[/tex]
Now put all the given values in this formula, we get:
[tex]pH+7.1=14.5[/tex]
[tex]pH=7.4[/tex]
Therefore, the pH of the solution is, 7.4
A solution has a pOH of 7.1 and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.
What is pH?pH is a figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values are more alkaline.
A solution has a pOH of 7.1 at 10 °C. We can calculate the pH at this temperature using the following expression.
pH + pOH = pKw
pH = pKw - pOH
pH = -log (2.93 × 10⁻¹⁵) - 7.1 = 7.4
where,
pH = -log [H⁺].pOH = -log [OH⁻].Kw is the ionic product of water.A solution has a pOH of 7.1 and a pH of 7.4 at 10 °C, being Kw = 2.93 × 10⁻¹⁵ at that temperature.
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The Bronsted-Lowry definition of acids and bases refers to the transfer of a "proton" from the acid to the base; however, the symbol for a proton (p+) is not generally used in this context. What is the chemical symbol that is commonly used to represent a "proton" in the context of Bronsted-Lowry acids and bases?
Answer: [tex]H^+[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
Example: For the given chemical equation:
[tex]H_3BO_3(aq.)+HS^-(aq.)\rightarrow H_2BO_3^-(aq.)+H_2S(aq.)[/tex]
Here, [tex]H_3BO_3[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]H_2BO_3^-[/tex] which is a conjugate base.
And, [tex]HS^-[/tex] is gaining a proton, thus it is considered as a base and after gaining a proton, it forms [tex]H_2S[/tex] which is a conjugate acid.
Thus the chemical symbol that is commonly used to represent a "proton" in the context of Bronsted-Lowry acids and bases is [tex]H^+[/tex]
Final answer:
In Brønsted-Lowry acid-base theory, a "proton" is symbolized by H₊ and is the key player in reactions where a proton donor (acid) transfers a proton to a proton acceptor (base).
Explanation:
The chemical symbol commonly used to represent a "proton" in the context of Brønsted-Lowry acids and bases is H₊. A proton essentially refers to a hydrogen ion (H₊) devoid of its electron, leaving just a single positive charge. In Brønsted-Lowry theory, a proton donor (acid) transfers a proton to a proton acceptor (base), thereby defining the roles of the acid and base in an acid-base reaction. This concept is crucial in understanding reactions such as the combination of ammonia and water, where NH₃ acts as the proton acceptor and water acts as the proton donor, forming NH₊₄ and OH₋.
As part of her nuclear stress test, Simone starts to walk on the treadmill. When she reaches the maximum level, Paul injects a radioactive dye with201Tl with an activity of 72 MBq . The radiation emitted from areas of her heart is detected by a scanner and produces images of her heart muscle. After Simone rests for 3 h, Paul injects more dye with 201Tl and she is placed under the scanner again. A second set of images of her heart muscle at rest is taken. When Simone's doctor reviews her scans, he assures her that she had normal blood flow to her heart muscle, both at rest and under stress.
Part A
If the half-life of 201Tl is 3.0 days, what is its activity, in megabecquerels after 3.0 days?
Express the activity to two significant figures and include the appropriate units.
Part B
What is its activity in megabecquerels after 6.0 days?
Express the activity to two significant figures and include the appropriate units.
Answer:
Part A: 36 MBq; Part B: 18 MBq
Explanation:
The half-life is the time it takes for half the substance to disappear.
The activity decreases by half every half-life
A =Ao(½)^n, where n is the number of half-lives.
Part A
3.0 da = 1 half-life
A = Ao(½) = ½ × 72 MBq = 36 MBq
Part B
6.0 da = 2 half-lives
A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq
Final answer:
Part A: The activity of 201Tl after 3.0 days is 36 MBq. Part B: The activity of 201Tl after 6.0 days is 18 MBq.
Explanation:
Part A: To calculate the activity of 201Tl after 3.0 days, we need to determine the number of half-lives that have passed. Since the half-life of 201Tl is 3.0 days, dividing 3.0 days by the half-life gives us 1 half-life. Each half-life reduces the activity of the isotope by half, so after 1 half-life, the activity would be half of the initial activity. Therefore, the activity after 3.0 days would be 72 MBq / 2 = 36 MBq.
Part B: To calculate the activity of 201Tl after 6.0 days, we can use the same method as in Part A. Since the half-life is 3.0 days, 6.0 days is equivalent to 2 half-lives. Each half-life reduces the activity by half, so after 2 half-lives, the activity would be 72 MBq / 2 / 2 = 18 MBq.
Analysis of an unknown sample indicated the sample contained 0.140 grams of N and 0.320 grams of O. The molecular mass of the compound was determined to be 92.02 amu. What is the molecular formula of the compound?
Answer:
the molecular formula of the compound is N2O4
Explanation:
- Find the empirical formula
mole of N present = mass of N divided by molar mass of N = 0.140/14 = 0.01 mole
mole of O present = mass of O divided by molar mass of O = 0.320/16 = 0.02 mole
Divide both by the smallest number of mole to determine the coefficient of each, the smallest number of mole is 0.01 thus:
quantity of N = 0.01/0.01 = 1
quantity of O = 0.02/0.01= 2
thus the empirical formula = NO2
- Now determine the molecular formula by finding the ratio of molecular formula and empirical formula
Molar mass of molecular formula = 92.02 amu = 92.02 g/mole
Molar mass of empirical formula NO2 = (14 + (16 x 2)) = 46 g/mole
the x factor = 92.02/46 = 2
Molecular formula = 2 x NO2 = N2O4
Final answer:
The molecular formula of the compound with 0.140 grams of N and 0.320 grams of O and a molecular mass of 92.02 amu is N₂O₄.
Explanation:
To determine the molecular formula of the compound, we first need to calculate the number of moles of nitrogen (N) and oxygen (O) in the sample. Using the atomic masses of N (14.0 amu) and O (16.0 amu), we divide the mass of each element in the sample by its respective atomic mass to find the moles:
Moles of O = 0.320 g / 16.0 g/mol = 0.02 mol
Next, we determine the simplest whole number ratio of the moles of each element, which gives us the empirical formula. In this case, the ratio of N to O is 1:2, so the empirical formula is NO₂.
Since we know the molecular mass of the compound is 92.02 amu, we can compare it to the mass of the empirical formula to find the molecular formula. The mass of NO₂ is 14.0 + (2 x 16.0) = 46.0 amu. Thus, the molecular mass of our compound (92.02 amu) is approximately twice that of NO₂, so the molecular formula is N₂O₄.
A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of the 50% solution and how many liters of the 90% solution will be used?
Answer:
50 ltr 150 ltr
Explanation:
this problem can be solved by the mixture and allegation concept which can be clearly understand from bellow figure in which the concentration of solution 1 is 50% and concentration of solution 2 is 90% before mixing after mixing with help bellow concept the ratio of concentration become 10:30
ratio of solution 1 and solution 2 =10:30
=1:3
total mixture is 200 liters
part of solution 1=[tex]\frac{1}{4}[/tex] ×200
=50 liters
part of solution 2=[tex]\frac{3}{4}[/tex] ×200
=150 liters
Consider the following reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is excess NO2 present. The molar masses are as follows: H2O = 18.02 g/mol, HNO3 = 63.02 g/mol. NO2(g) + H2O(l) → HNO3(aq) + NO(g)
Answer: The mass of water required will be 10.848 g.
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)
For nitric acid:Given mass of nitric acid = 75.9 g
Molar mass of nitric acid = 63 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of nitric acid}=\frac{75.9g}{63g/mol}=1.204mol[/tex]
For the given chemical equation:
[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)[/tex]
By Stoichiometry of the reaction:
2 moles of nitric acid is produced by 1 mole of water.
So, 1.204 moles of nitric acid will be produced by = [tex]\frac{1}{2}\times 1.204=0.602mol[/tex] of water.
Now, calculating the amount of water, we use equation 1:
Moles of water = 0.602 mol
Molar mass of water = 18.02 g/mol
Putting values in equation 1, we get:
[tex]0.602=\frac{\text{Mass of water}}{18.02g/mol}\\\\\text{Mass of water}=10.848g[/tex]
Hence, the mass of water required will be 10.848 g.
To form 75.9 g of HNO3, 10.85 g of water is required.
Explanation:To determine the number of grams of water required to form 75.9 g of HNO3, we need to use the given balanced equation and apply stoichiometry. From the equation, we can see that 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3. Using the molar masses of H2O and HNO3, we can calculate the amount of water needed.
First, convert the mass of HNO3 to moles using its molar mass:
HNO3: 75.9 g ÷ 63.02 g/mol = 1.204 mol HNO3
Next, use the stoichiometric ratio to find the moles of water:
H2O: (1.204 mol HNO3) × (1 mol H2O) ÷ (2 mol HNO3) = 0.602 mol H2O
Finally, convert the moles of water to grams using its molar mass:
0.602 mol H2O × 18.02 g/mol = 10.85 g H2O
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Estimate the vapor pressure of methanol at 25°C. The heat of vaporization for methanol is 37,400 J/mol and the boiling point is 65°C
The vapor pressure of methanol at 25°C can be estimated using the Clausius-Clapeyron equation. By plugging in the values for methanol's boiling point, enthalpy of vaporization, and the desired temperature, we can solve for the vapor pressure. The estimated vapor pressure of methanol at 25°C is approximately 0.147 atm.
Explanation:In order to estimate the vapor pressure of methanol at 25°C, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure, temperature, and enthalpy of vaporization. The equation is:
ln(P2/P1) = -ΔHvap/R × (1/T2 - 1/T1)
Where P1 is the known vapor pressure at a known temperature (T1), P2 is the vapor pressure at the desired temperature (T2), ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. Plugging in the values for methanol:
P1 = vapor pressure at boiling point = 1 atm, T1 = boiling point = 65°C = 338 K, T2 = desired temperature = 25°C = 298 K, ΔHvap = 37,400 J/mol, R = 8.314 J/(mol·K)
the equation becomes:
ln(P2/1) = -37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K)
Solving for P2:
P2 = 1 × e ^ (-37,400 J/mol / (8.314 J/(mol·K)) × (1/298 K - 1/338 K))
P2 ≈ 0.147 atm
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Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [P4O10]eq = 2.000 moles, [P4]eq = 3.000 moles, [O2]eq = 4.000 M P4O10(s) ↔ P4(s) + 5 O2(g)
Answer : The value of [tex]K_c[/tex] for the following reaction will be, 1024
Explanation :
[tex]K_c[/tex] is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants
The given balanced equilibrium reaction is,
[tex]P_4O_{10}(s)\rightleftharpoons P_4(s)+5O_2(g)[/tex]
As we know that the concentrations of pure solids are constant that means they do not change. Thus, they are not included in the equilibrium expression.
The expression for equilibrium constant for this reaction will be,
[tex]K_c=[O_2]^5[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=(4.000)^5[/tex]
[tex]K_c=1024[/tex]
Therefore, the value of [tex]K_c[/tex] for the following reaction will be, 1024
The value of the equilibrium constant for the given reaction is 1,024.
How we calculate the equilibrium constant?
Equilibrium constant of any chemical reaction at the equilibrium state is calculated as:
Equilibrium constant Kc = Concentration of product / Concentration of reactant
Given chemical reaction is:
P₄O₁₀(s) → P₄(s) + 5O₂(g)
Equilibrium constant for the above reaction is written as:
Kc = [P₄][O₂]⁵ / [P₄O₁₀]
For this reaction only concentration of oxygen gas is taken into consideration, as all other quantities are present in solid form and their value at equilibrium is 1. So, value of Kc is calculated as:
Kc = [O₂]⁵
Given concentration of O₂ = 4M
Kc = (4)⁵ = 1,024
Hence, 1,024 is the value of Kc.
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For a stationary fluid, the pressure will vary in the x, y and z directions.T/F
Answer:
True
Explanation:
For a stationary fluid, the pressure will vary in the x, y and z directions.
This statement is true.
Which group 1 element exhibits slightly different chemistry from the others?
Li
K
Rb
Na
Cs
Answer:
Li
Explanation:
The group 1 elements are the alkali elements, and they are the ones that have one valence electron int heir outer shell, along wiht them Hydrogen is also an element that has one valence electron but is not considered group 1 because it is not a solid in temperature room, but a gas and has different properties, from the options the most different from the others is litium, because it can react with the same elements than the other elements in group one, plus nitrogen in a very violent way.
The group 1 element that exhibits slightly different chemistry from the others is: Lithium (Li).
The chemical elements found in group 1 of the periodic table are highly electro positive metals and have a single (1) valence electron in its outermost shell.
Some examples of these chemical elements (alkali metals) are;
Sodium (Na).Hydrogen (H).Potassium (K).Lithium (Li).Li is the symbol for the chemical element referred to as Lithium.
Lithium (Li) is an alkali metal and as such it is found in group 1 of the periodic table with a single (1) valence electron in its outermost shell.
Also, the electronic configuration of Lithium (Li) is written as;
1s²2s¹As a result of the small size of Lithium (Li), it is exhibits slightly different chemistry from the others such as:
High melting and boiling point.Harder than other alkali metals.Less reactive with water and oxygen.High ionization energy.Read more: https://brainly.com/question/20346057
The pKa values for the carboxy and ammonium protons of phenylalanine are 2.58 and 9.24, respectively. What is the isoelectric point of phenylalanine? Draw the structure of phenylalanine at its isoelectric point.
Answer:The isoelctric point of phenylalanine is 5.91. Kindly refer the attachment for structure.
Explanation:
Isoeectronic point is the pH at which the amino acid do not migrate in presence of an electric field.Hence at isoelectric point the molecule exists as a neutral molecule.
So for a molecule to not migrate in electric field it must be neutral. At isoelectric point the net charge on the amino acid molecule is zero.
The amino acid at isoelectric point exists as a zwitter ion which has a equal positive and equal negative charge making the net charge of the amino acid to be zero.
since the amino acid molecule has two kinds of functional group present which are an amine group and a carboxylic acid group.
So to calculate the isoelectric point we must know the pKa of these functional groups.
The formula for calculating the isoelectic point (pI) is :
[tex]pI=[pKa1+pKa2]\times 1/2[/tex]
pI=Isoelectric point
pKa1=pKa of carboxy group
pKa2=pKa of amine group
we have the pKa1 and pKa2 given in the question so putting these values in above formula we can calculate the isoelectric point.
pKa1=2.58
pKa2=9.24
pI=1/2×[pKa1+pKa2]
pI=1/2×[2.58+9.24]
pI=5.91
The isoelectric point is 5.91 hence at a pH of 5.91 phenylalanine would exist as a zwitterion and would not migrate in presence of a electric field.
Determine the calcium carbonate equivalent (CCE) of the following compounds: (amount that has the same neutralizing value as 100 g pure CaCO3) (a) MgO (b) Mg(OH)2 (c) and CaMg(CO3)2.
Answer:
a) 40 g
b) 58 g
c) 184 g
Explanation:
The calcium carbonate equivalent of any compound is calculated by calculating it molar mass.
The molar mass of the compound is the calcium carbonate equivalent as it corresponds to 1 mole of the compound and equals to one mole of calcium carbonate then.
a) MgO
Atomic mass of Mg = 24
Atomic mass of O = 16
Molar mass = CCE = 24+16 = 40 g
b) Mg(OH)₂
Atomic mass of Mg = 24
Atomic mass of O = 16
Atomic mass of H = 1
Molar mass of Mg(OH)₂= CCE = 24 + (2X16)+2(X1) = 58g
c) CaMg(CO₃)2
Atomic mass of Mg = 24
Atomic mass of O = 16
Atomic mass of C = 12
Atomic mass of Ca = 40
Molar mass = CCE = 40 + 24 + (2X12) + (6X16) = 184 g.
Based on Coulomb's Law of electrostatic attraction of oppositely-charged species, which of the following chlorine-containing compounds would be predicted to have the greatest (most exothermic) lattice energy? CaCl2 NaCl MgCl2 KCl CCl4
Based on Coulomb's Law of electrostatic attraction of oppositely-charged species, the chlorine-containing compound that is predicted to have the greatest (most exothermic) lattice energy is; MgCl₂
The lattice energy is defined as the energy required to dissociate one mole of an ionic compound to its constituent gaseous ions.
Coulomb's law of electrostatic attraction states that the force of attraction (F) between two oppositely charged particles is directly proportional to the product of the charges of the particles ( q₁ and q ₂ ) and inversely proportional to the square of the distance between the particles.
Thus; F = (q₁ × q ₂)/r²
Now, Lattice energy is inversely proportional to the size of the ions. This implies that as the size of the ions increases, then the lattice energy will decrease. The size of the ions is also called the atomic radii.
Thus, as atomic radii increases, lattice energy will decrease.
From the formula above and definition, we can tell that Lattice energy decreases down a group. Also, lattice energy will increase as the charges increase.
Now, the compounds we are dealing with are;
CaCl₂ - Calcium Chloride
NaCl - Sodium Chloride
MgCl₂ - Magnesium Chloride
KCl - Potassium Chloride
CCl₄ - Carbon tetrachloride
The charges of the metals that form the chlorides above are;
Ca = +2
Na = +1
Mg = +2
K = +1
The highest charges are Ca and Mg.
Now, they both belong to group two of the periodic table with Ca below Mg in the periodic table and as such Mg will have a greater Lattice energy.
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From the list of chlorine-containing compounds, MgCl₂ would have the greatest lattice energy. Mg²⁺ has a greater charge than the other cations, resulting in stronger electrostatic attraction and most exothermic energy. CCl₄ is a covalently bonded molecule not an ionic compound, so does not possess lattice energy.
Explanation:Based on Coulomb's Law of electrostatic attraction, the strongest (most exothermic) lattice energy would be present in the compound with the highest positively charged ion and smallest ionic radii, which results in a significant increase in lattice energy. This comes from the idea that lattice energy increases with higher charges and decreases with larger ionic radii.
In this case, it would be MgCl₂, because Mg⁺ has a greater charge than the other cations listed (Ca²⁺, Na⁺, K⁺). Greater charge results in stronger electrostatic attraction and hence the most exothermic lattice energy.
It's worth noting that CCl₄ is a covalently bonded molecule and not an ionic compound, thus it does not have lattice energy in the context of ionic compounds.
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You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2HPO4)and you need 1L of the buffer at pH 7.00 with a total phosphate concentration(NaH2PO4+ Na2HPO4) of 0.100 M. Hint:Phosphoric acid (H3PO4), a triprotic acid, has 3 pKa values: 2.14, 6.86, and12.4. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4;138g/mol) and disodium hydrogen phosphate (Na2HPO4;142g/mol) needed to prepare this buffer.
Answer:
For disodium hydrogen phosphate:
5.32g Na2HPO4
For sodium dihydrogen phosphate:
7.65g Na2HPO4
Explanation:
First, you have to put all the data from the problem that you going to use:
-NaH2PO4 (weak acid)
-Na2HPO4 (a weak base)
-Volume = 1L
-Buffer pH = 7.00
-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M
What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:
H3PO4 = H2PO4 + H+ pKa1 = 2.14
H2PO4 = HPO4 + H+ pKa2 = 6.86
HPO4 = PO4 + H+ pKa3 = 12.4
So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:
pKa2= 6.86
Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:
pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]
The solution of the problem is attached to this answer.
To prepare 1L of a phosphate buffer at pH 7 with a total concentration of 0.1 M, one would need to mix equal molar amounts of NaH2PO4 and Na2HPO4, resulting in 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4.
The relevant pKa value for the pH target of 7.00 is 6.86, which is close to the second pKa of phosphoric acid. The Henderson-Hasselbalch equation is as follows:
pH = pKa + log([A-]/[HA]), where:
At pH 7.00, the ratio [A-]/[HA] is 1:1 because pH = pKa. Thus, we need equal molar amounts of NaH2PO4 and Na2HPO4. Since the total molarity is 0.1 M, this means we need 0.05 M of each component.
Therefore, you would need to weigh 6.9 g of NaH2PO4 and 7.1 g of Na2HPO4 to prepare your buffer.
In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer.
Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 505 liters per second of dinitrogen are consumed when the reaction is run at 172 oC and 0.88 atm. Calculate the rate at which ammonia is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.
Answer:
The rate at which ammonia is being produced is 0.41 kg/sec.
Explanation:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex] Haber reaction
Volume of dinitrogen consumed in a second = 505 L
Temperature at which reaction is carried out,T= 172°C = 445.15 K
Pressure at which reaction is carried out, P = 0.88 atm
Let the moles of dinitrogen be n.
Using an Ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol[/tex]
According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.
Then 12.1597 mol of dinitrogen will produce :
[tex]\frac{2}{1}\times 12.1597 mol=24.3194 mol[/tex] of ammonia
Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol
=413.43 g=0.41343 kg ≈ 0.41 kg
505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.
Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation C6H12O6(aq)+6O2(g)⟶6CO2(g)+6H2O(l) Calculate the number of grams of oxygen required to convert 48.0 g of glucose to CO2 and H2O. mass of O2: g Calculate the number of grams of CO2 produced. mass of CO2: g
Answer:
70.3824 grams of carbondioxide is produced.
Explanation:
[tex]C_6H_{12}O_6(aq)+6O_2(g)⟶6CO_2(g)+6H_2O(l)[/tex]
Moles of glucose = [tex]\frac{48.0 g}{180 g/mol}=0.2666 mol[/tex]
According to reaction 1 mole of glucose reacts with 6 mole of oxygen gas.
Then 0.2666 mol of gluocse will reactwith :
[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of oxygen
Mass of 1.5996 moles of oxygen gas:
[tex]1.5996 mol\times 32 g/mol = 51.1872 g[/tex]
51.1872 grams of oxygen are required to convert 48.0 grams of glucose to [tex]CO_2[/tex] and [tex]H_2O[/tex].
According to reaction, 1 mol, of glucose gives 6 moles of carbon dioxide.
Then 0.2666 moles of glucose will give:
[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of carbon dioxide
Mass of 1.5996 moles of carbon dioxide gas:
[tex]1.5996 mol\times 44 g/mol = 70.3824 g[/tex]
70.3824 grams of carbondioxide is produced.
Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.
Let's consider the overall equation for the combustion of glucose in the human body.
C₆H₁₂O₆(aq) + 6 O₂(g) ⟶ 6 CO₂(g) + 6 H₂O(l)
We can calculate the grams of oxygen required to react with 48.0 g of glucose considering the following relations.
The molar mass of glucose is 180.16 g/mol.The molar ratio of glucose to oxygen is 1:6.The molar mass of oxygen is 32.00 g/mol.[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molO_2}{1molGlucose} \times \frac{32.00 gO_2}{1 molO_2} = 51.2 gO_2[/tex]
We can calculate the grams of carbon dioxide produced from 48.0 g of glucose considering the following relations.
The molar mass of glucose is 180.16 g/mol.The molar ratio of glucose to carbon dioxide is 1:6.The molar mass of carbon dioxide is 44.01 g/mol.[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molCO_2}{1molGlucose} \times \frac{44.01 gCO_2}{1 molCO_2} = 70.4 gCO_2[/tex]
Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.
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According to Raoult's law, which statement is FALSE? a) The vapor pressure of a solvent over a solution decreases as its mole fraction increases. b) The vapor pressure of a solvent over a solution is less than that of pure solvent. c) The greater the pressure of a gas over a solution the greater its solubility. d) The solubility of a gas increases as the temperature decreases. e) Ionic solutes dissociate in solution causing an enhancement of all colligative properties.
Answer:
a) The vapor pressure of a solvent over a solution decreases as its mole fraction increases.
According to Raoult's law, the statement which is FALSE is:
A) The vapor pressure of a solvent over a solution decreases as its mole fraction increases.Raoult's law states that if the pressure of the content of a liquid remains constant to the temperature, then it is proportional to the mole fraction of the mixture.
As a result of this, the vapor pressure of a solvent over a solution is less than the pure solvent and the greater the pressure of a gas over a solution, the greater the solubility.
Therefore, the correct answer is option A because it is false.
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Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: 4H2 O(g) + 3Fe(s) ⟶ Fe3 O4 (s) + 4H2 (g) (a) Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O? (b) Answer the question.
Answer : The volume of [tex]H_2[/tex] will be, 0.2690 L
Solution :
(a) Steps involved for this problem are :
First we have to calculate the moles of [tex]H_2O[/tex].
Now we have to calculate the volume of hydrogen gas by using the ideal gas equation.
(b) First we have to calculate the moles of [tex]H_2O[/tex].
[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{15g}{18g/mole}=0.833moles[/tex]
The balanced chemical reaction is,
[tex]4H_2O(g)+3Fe(s)\rightarrow Fe_3O_4(s)+4H_2(g)[/tex]
From the balanced chemical reaction, we conclude that the moles of hydrogen is equal to the moles of water.
Thus, the moles of hydrogen gas = 0.833 mole
Now we have to calculate the volume of hydrogen gas.
Using ideal gas equation,
[tex]PV=nRT[/tex]
where,
n = number of moles of gas = 0.833 mole
P = pressure of the gas = [tex]745torr=\frac{745}{760}=0.98atm[/tex]
conversion used : (1 atm = 760 torr)
T = temperature of the gas = [tex]20^oC=273+20=293K[/tex]
R = gas constant = 0.0821 Latm/moleK
V = volume of gas = ?
Now put all the given values in the above equation, we get :
[tex](0.98atm)\times V=(0.833mole)\times (0.0821Latm/moleK)\times (293K)[/tex]
[tex]V=0.2690L[/tex]
Therefore, the volume of [tex]H_2[/tex] will be, 0.2690 L
Answer:
V = 20.4 L
Explanation:
Step 1: Write the balanced equation
4 H₂O(g) + 3 Fe(s) ⟶ Fe₃O₄(s) + 4 H₂(g)
Step 2: Find the moles of H₂
We can establish the following relations.
The molar mass of H₂O is 18.0 g/molThe molar ratio of H₂O to H₂ is 4:4.The moles of H₂ obtained from 15.0 g of H₂O is:
[tex]15.0gH_{2}O.\frac{1molH_{2}O}{18.0gH_{2}O} .\frac{4molH_{2}}{4molH_{2}O} =0.833molH_{2}[/tex]
Step 3: Find the volume of H₂
We will use the ideal gas equation.
P = 745 torr × (1 atm/760 torr) = 0.980 atm
V = ?
n = 0.833 mol
R = 0.08206 atm.L/mol.K
T = 20°C + 273 = 293 K
P × V = n × R × T
0.980 atm × V = 0.833 mol × (0.08206 atm.L/mol.K) × 293 K
V = 20.4 L
Consider the reaction: HCN (aq) + H2O (l) ⇄ CN- (aq) + H3O + (aq) Which of the following statements will decrease the amount of work the system could perform? (a) Boil off water from the container (increasing the concentration of all species) (b) Add solid NaOH to the reaction (assume no volume change) (c) Selectively remove CNfrom the solution (d) Add water to the reaction vessel (e) Increase the concentration of the HCN
Answer: Option (b) is the correct answer.
Explanation:
The given chemical reaction shows that hydrogen cyanide acid has been added to water which results in the formation of hydronium ion and cyanide ion.
Also, when we add a base like sodium hydroxide (NaOH) to HCN then it will help in accepting a proton ([tex]H^{+}[/tex]) from hydrogen cyanide. As a result, formation of [tex]CN^{-}[/tex] anion will be rapid and easy.
This will make the system not to do any extra work. So, amount of work done by system will decrease.
Thus, we can conclude that out of the given options, add solid NaOH to the reaction (assume no volume change) will decrease the amount of work the system could perform.
Boiling off water from the container will increase the concentration of all species and shift the equilibrium toward the reactants, thus decreasing the potential work the system could perform, aligning with Le Chatelier's Principle. Option A is correct.
Which, action would decrease the amount of work a system, involving the equilibrium reaction of HCN (aq) with water to produce CN⁻ (aq) and H₃O⁺ (aq), could perform. Various actions can shift the equilibrium of this reaction, affecting the system's potential work. According to Le Chatelier's Principle, the system will respond to minimize the effect of any change.
Boiling off water ((a)) from the container will increase the concentration of all species, shifting the equilibrium to the left, thus diminishing the potential work by favoring reactants.Adding solid NaOH ((b)) will increase the hydroxide ion concentration, which consumes H₃O⁺ ions, driving the reaction to the right and potentially increasing work, contradicting the asked condition.Removing CN- ((c)) selectively from the solution will drive the reaction to the right, to replenish CN⁻ ions, potentially increasing the system's capacity to do work, which is against the question's condition.Adding water ((d)) to the reaction vessel will dilute all reactive species, shifting the equilibrium towards CN⁻ and H₃O⁺ formation, which doesn't directly answer the question of decreasing work.Increasing the concentration of HCN ((e)) will push the equilibrium to the right, thus increasing the amount of work the system could perform by forming more products.The action that will decrease the amount of work the system could perform is (a) boiling off water from the container, as it favors the formation of reactants over products.
Hence, A. is the correct option.
The element chlorine (Cl) has two isotopes: chlorine‑35 and chlorine‑37. Approximately 75.5% of chlorine atoms have 18 neutrons and 17 protons, and the other 24.5% have 20 neutrons and 17 protons. Using the isotopic composition provided, calculate the average atomic mass of chlorine. Round your answer to the tenths place.
Answer:
35.5 amu is the average atomic mass of chlorine
Explanation:
Fractional abundance of chlorine‑35 = 75.5%
Fractional abundance of chlorine‑37 = 24.5%
Average atomic mass is equal to summation of products of all isotopes masses into their fractional abundance.
Average atomic mass = Σ(Mass of an isotope × fractional abundance)
Average atomic of chlorine :
[tex]35 amu\times 0.755+37 amu\times 0.245=35.49 amu\approx 35.5 amu[/tex]
35.5 amu is the average atomic mass of chlorine
Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the thermal decomposition of calcium carbonate: CaCO3(s) → CaO(s) CO2(g) Calculate the yearly release of CO2 (in kg) to the atmosphere if the annual production of CaO in the United States is 8.6 × 1010 kg.
The question involves calculating the CO2 emissions from the annual production of calcium oxide in the U.S., based on the stoichiometry of the reaction converting CaCO3 to CaO, which releases CO2.
Explanation:The question is about calculating the yearly release of CO2 to the atmosphere resulting from the annual production of calcium oxide (CaO) in the United States, given that the reaction for producing CaO from calcium carbonate (CaCO3) releases CO2. Since the reaction is CaCO3(s) → CaO(s) + CO2(g), for each mole of CaCO3 decomposed, one mole of CO2 is released. Calcium carbonate (mol weight = 100.09 g/mol) decomposes to give calcium oxide (mol weight = 56.08 g/mol) and carbon dioxide (mol weight = 44.01 g/mol). Given the annual production of CaO is 8.6 × 10¹° kg, we first convert this mass to moles (using CaO's molar mass), and then calculate the corresponding moles (and mass) of CO2 released.
What concentration of Ni2+ ion remains in solution after electrolysis of 100. mL of 0.250 M NiSO4 solution when using a current of 2.40 amperes for 30.0 minutes? Assume Ni metal is plated out.
Answer:
the concentration remains = 0.138 M
Explanation:
Let us calculate the charge passed:
The charge passed during the electrolysis = current X time(s)
charge passed = 2.40X30X60= 2160 C
we know that
96485 C = 1 F
Therefore 2160 C = 0.0224 F
according to Faraday's law of electrolysis if we pass one Faraday of charge through an electrolytic solution, 1 gram equivalent of metal will be deposited.
the gram equivalent of nickel deposited = 0.0224
The moles of Nickel deposited = 0.0224/2 = 0.0112 mol
the initial moles of Nickel ions present in solution is
[tex]moles=molarityXvolume(L)=0.250X0.1=0.0250mol[/tex]
The moles of nickel ion gets consumed = 0.0112
So moles of nickel ion left after electrolysis = 0.0138
[tex][Ni^{+2}]=\frac{moles}{volume(L)}\frac{0.0138}{0.1}=0.138M[/tex]
The concentration of Ni2+ ions remaining in the solution can be found by applying Faraday's first law of electrolysis and subtracting the moles of Ni2+ discharged from the initial moles present in the solution.
Explanation:The electrolysis process involves a chemical reaction in which electricity is used to break down a substance into its constituent elements. In this scenario, we are determining the concentration of Ni2+ in the solution after the electrolysis process of NiSO4.
As per Faraday's first law of electrolysis, the mass of any element discharged during electrolysis is directly proportional to the quantity of electricity (current) passed. Thus, using Faraday's constant (96485 C/mol e-) and the given current and time, we can calculate how many moles of Ni2+ are plated out.
The number of moles of Ni2+ discharged would be equal to the number of moles initially present in the 100.0 mL of 0.250 M NiSO4 solution. By subtracting the moles of Ni2+ that were discharged from the total initial moles of Ni2+, we can obtain the number of moles of Ni2+ remaining in the solution. This value can be converted back into molarity by dividing the moles of Ni2+ remaining by the volume of the solution in liters.
In this way, we can find the concentration of Ni2+ ions that remain in the solution after electrolysis.Learn more about Electrolysis here:https://brainly.com/question/33301115
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Ritonavir oral solution contains, in addition to ritonavir, 43.2% alcohol and 26.57% propylene glycol. Calculate the quantity of alcohol in a 240 mL bottle of the oral solution. (Round to two decimal places.)
Hey there!:
240 mL bottle of the oral solution , so 43.2% alcohol
Therefore:
240 mL * 43.2 / 100
= 2.4 * 43.2
=> 103.68 mL of alchohol
Hope this helps!
Ritonavir oral solution contains, in addition to ritonavir, 43.2% alcohol and 26.57% propylene glycol. The quantity of alcohol in a 240 ml bottle of the oral solution is 103.68 ml of alcohol.
What is alcohol?Alcohol is a drink that is made up of rotten fruits and vegetables, It is made by fermentation. They cause unconsciousness to the brain and the body. It creates hallucinations.
Ritonavir is taken with meals twice a day. Ask your doctor or pharmacist to explain any instructions on your prescription label that you are unsure about following.
Given the 43.2% alcohol and 26.57% propylene glycol. The bottle is of 240 ml.
240 mL x 43.2 / 100
2.4 x 43.2 = 103.68 ml of alcohol.
Thus, the quantity of alcohol in a 240 ml bottle is 103.68 ml of alcohol.
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True or False Materials that have predominantly ionic or covalent bonds have many free electrons.
Answer:
False
Explanation:
Atomic bonding can be of two types:
Ionic bonds: The bond formed between two opposite charged ions. The strong electrostatic force acting between the ions make them compounds of high stability . Example: NaCl
Covalent bonds: The bond formed by sharing of the electrons between the atoms. The balance between the attractive force (nucleus, electron) and repulsive force (electron, electron or nucleus, nucleus) makes the compound stable. Example: [tex]CH_4[/tex]
The bonds are either formed by transferring (losing and gaining) or sharing of electrons. Thus, compound forming these bonds share or transfer electrons to attain noble gas configuration. Thus, they lack free electrons.
A rigid tank is divided into two equal volumes. One side contains 2 kmol of nitrogen N2 at 500 kPa while the other side contains 8 kmol of CO2 at 200 kPa. The two sides are now connected and the gases are mixed and forming a homogeneous mixture at 250 kPa. Find the partial pressure of the CO2 in the final mixture.
Answer:
The partial pressure of the [tex]CO_2[/tex] in the final mixture is 200 kPa.
Explanation:
Pressure of nitrogen gas when the two tanks are disconnected = 500 kPa
Pressure of the carbon-dioxide gas when the two tanks are disconnected = 200 kPa
Moles of nitrogen gas =[tex]n_1= 2 kmol[/tex]
Moles of carbon dioxide gas =[tex]n_2=8 kmol[/tex]
After connecting both the tanks:
The total pressure of the both gasses in the tank = p = 250 kPa
According to Dalton' law of partial pressure:
Total pressure is equal to sum of partial pressures of all the gases
Partial pressure of nitrogen =[tex]p_{N_2}^o[/tex]
Partial pressure of carbon dioxide=[tex]p_{CO_2}^o[/tex]
[tex]p_{N_2}^o=p\times \frac{n_1}{n_1+n_2}[/tex]
[tex]p_{N_2}^o=250 kPa\times \frac{0.2}{0.2+0.8}=50 kPa[/tex]
[tex]p_{CO_2}^o=p\times \frac{n_2}{n_1+n_2}[/tex]
[tex]p_{CO_2}^o=250 kPa\times \frac{0.8}{0.2+0.8}=200 kPa[/tex]
The partial pressure of the [tex]CO_2[/tex] in the final mixture is 200 kPa.
Which of the following four statements is/are accurate with respect to glycolysis?
A. Glycolysis involves the conversion of monosaccharides into glucose.
B. Glycolysis involves the breakdown of glucose into pyruvate.
C. Glycolysis involves the conversion of pyruvate into glucose to glyceraldehyde-3-phosphate.
D. Glycolysis involves the conversion of pyruvate into acetyl CoA.
Answer:
B. Glycolysis involves the breakdown of glucose into pyruvate.
Explanation:
Glycolysis is the biological process whereby one glucose molecule is broken down to two pyruvate molecules. There are a lot of enzymatic processes that are involved in this. It is one of the most important reactions in the world because it allows living cells to harness chemical energy from organic molecules
You are holding a container of 28.7L of dinitrogen tetroxide. How many grams of gas are inside? Step by step.
Answer:
=71.76 grams
Explanation:
At Room temperature and pressure, 1 mole of an ideal gas occupies a volume of 24 liters.
Therefore, the number of moles occupied by 28.7 L is:
(28.7×1)/24=1.196 moles.
Mass=Number of moles× RMM
RMM of N₂O₂ is 60
mass=1.196 moles× 60 grams/mol
=71.76 grams
A saturated solution of magnesium fluoride, MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 1.18×10−3 M . Calculate Ksp for MgF2.
Hey there!:
Ksp =(Mg⁺²)(2F⁻)²
Ksp = (1.18*10⁻³)(2*1.18*10⁻³)²
Ksp = 6.57*10⁻⁹
Hope that helps!
Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 27.32 °C are mixed. The specific heat of water is 4.184 J/gC). Record your answer in scientific notation using 3 significant figures.
Answer:
The final temperature of the water system is 51.5 °C.
Explanation:
A 50.0 gram sample of water initially at 100 °C
Mass of the water = 50 g
Initial temperature of the water = [tex]T_i=100[/tex]
Final temperature of the water after mixing = [tex]T_f=[/tex]
-Q' = heat lost by 50.0 g of water
[tex]-Q'=mc\Delta (T_f-T_i)[/tex]
A 100.0 gram sample of water initially at 27.32°C
Mass of the water ,m'= 10.00 g
Initial temperature of the water = [tex]T_i'=100[/tex]
Final temperature of the water after mixing = [tex]T_f=[/tex]
Q = heat gained by 100.0 g of water after mixing
[tex]Q=m'c\Delta (T_f-T_i')[/tex]
-Q'=Q (Energy remains conserved)
[tex]-(mc\Delta (T_f-T_i))=m'c\Delta (T_f-T_i')[/tex]
[tex]-(50 g\times 4.184J/g^oC(T_f-100^oC))=100 g\times 4.184J/g^oC(T_f-27.32^oC)[/tex]
[tex]T_f=51.54^oC\approx 51.5^oC[/tex]
The final temperature of the water system is 51.5 °C.
To determine the final temperature of the system, we consider the exchange of heat between the warm and cold water samples. We use the heat transfer formula and conservation of energy principle. Then, solving for 'T', we get the final temperature – in scientific notation.
Explanation:This question revolves around the concept of specific heat and its application in heat transfer. When two substances with different initial temperatures are mixed, they will exchange heat until an equilibrium temperature is reached.
The equation to calculate heat exchanged is: Q = mcΔT, where 'm' is mass, 'c' is specific heat, and 'ΔT' is the change in temperature.
In this context, water with a temperature of 100 °C will lose heat (Qhot), whereas water at 27.32 °C will gain heat (Qcold). When the system reaches thermal equilibrium, Qhot = -Qcold due to the law of conservation of energy.
So, we will have two equations:
Qhot = (50.0 g)(4.184 J/g°C)(100 - T)Qcold = (100.0 g)(4.184 J/g°C)(T - 27.32)We then solve for T, which represents the final temperature of the system.
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A tank has a volume of 0.1 m3 and is filled with He gas at a pressure of 5 x 106 Pa. A second tank has a volume of 0.15 m' and is filled with He gas at a pressure of 6 x 106 Pa. A valve connecting the two tanks is opened. Assumıng He to be a monatomic ideal gas and the walls of the tanks to be adiabatic and rigıd, find the final pressure of the system Hint: Note that the internal energy is constant.
Answer:
[tex]P=5.6*10^{6} Pa[/tex]
Explanation:
Consider that, as the system is adiabatic, [tex]U_{1}= U_{2}[/tex] where U1 and U2 are the internal energies before the process and after that respectively.
Consider that: [tex]U=H-PV[/tex], and that the internal energy of the first state is the sum of the internal energy of each tank.
So, [tex]H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}[/tex]
Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: [tex]H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}[/tex]
So, [tex]P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}[/tex]
Isolating [tex]P_{2}[/tex],
[tex]P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}[/tex]
[tex]V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}[/tex]
So,
[tex]P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa[/tex]