How many molecules of carbon dioxide, CO2, are in 38.5 moles?

True or False: A heterozygous genotype produces the dominant phenotype because the dominant allele masks the recessive allele.

jasenka [17]

This is true! Heterozygous would be Dd, and the dominant masks the recessive

4 0

3 years ago

The heat of combustion of propane, C3H8 (g) is -2057 kJ/mol. What would be the enthalpy change if enough propane was burned to g

Digiron [165]

Considering the reaction stoichiometry, the enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.

The balanced reaction is:

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

The heat of combustion of propane, C₃H₈, is -2057 kJ/mol. This is, 2057 kJ is released for every 1 mol C₃H₈.

So to determine the enthalpy change if enough propane was burned to emit 12 moles of carbon dioxide, you must take into account the stoichiometry of the reaction.

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Then you can apply the following rule of three: if by stoichiometry 3 moles of CO₂ are produced by 1 mole of C₃H₈, 12 moles of CO₂ are produced by how many moles of C₃H₈?

$amount of moles of C_{3} H_{8} =\frac{12 moles of CO_{2}x1 mole of C_{3} H_{8} }{3 moles of CO_{2}}$

amount of moles of C₃H₈= 4 moles

So to determine the enthalpy change, you can apply the following rule of three: If for each mole of C₃H₈ 2057 kJ are released, for 4 moles of C₃H₈ how much heat is released?

$Heat released=\frac{4 molesx2057 kJ}{1 mole}$

Heat released= 8228 kJ

The enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.

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5 0

3 years ago

A disadvantage of sexual reproduction is that

sineoko [7]

The disadvantage of sexual reproduction is that outside influences can determine the viability of the offspring. In humans, for example, a failure for a mother to consume an adequate amount of folic acid is directly linked to neural tube birth defects. This defect occurs at the earliest stages of development, often when a woman doesn’t know she is pregnant, which means the folic acid must be consumed when attempting to conceive.

8 0

3 years ago

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311

LuckyWell [14K]

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

5 0

4 years ago

Calculate the molecular weight of a dibasic acid.0.56gm of which is required 250ml of N/20 sodium hydroxide solution for neutral

Alekssandra [29.7K]

Answer: The molecular weight of the dibasic acid is 89.6 g/mol

Explanation:

Normality is defined as the amount of solute expressed in the number of gram equivalents present per liter of solution. The units of normality are eq/L. The formula used to calculate normality:

$\text{Normality}=\frac{\text{Given mass of solute}\times 1000}{\text{Equivalent mass of solute}\times \text{Volume of solution (mL)}}$ ....(1)

We are given:

Normality of solution = $\frac{1}{20}=0.05N$

Given mass of solute = 0.56 g

Volume of solution = 250 mL

Putting values in equation 1, we get:

$0.05=\frac{0.56\times 1000}{\text{Equivalent mass of solute}\times 250}\\\\\text{Equivalent mass of solute}=\frac{0.56\times 1000}{0.05\times 250}=44.8g/eq$

Equivalent weight of an acid is calculated by using the equation:

$\text{Equivalent weight}=\frac{\text{Molar mass}}{\text{Basicity}}$ .....(2)

Equivalent weight of acid = 44.8 g/eq

Basicity of an acid = 2 eq/mol

Putting values in equation 2, we get:

$44.8g/eq=\frac{\text{Molar mass}}{2eq/mol}\\\\\text{Molar mass}=(44.8g/eq\times 2eq/mol)=89.6g/mol$

Hence, the molecular weight of the dibasic acid is 89.6 g/mol

7 0

3 years ago