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3 years ago

What is indicated by 'A' in the figure below? A. -VE B. +Ea C. -Ea D. +VE

Chemistry

1 answer:

makkiz [27]3 years ago

6 0

Answer:

+VE

Explanation:

If we look at the reaction profile pictured in the question, we can easily identify A as the enthalpy of the reaction. The enthalpy of reaction (ΔHrxn) is usually defined as the difference between the total enthalpy (heat content) of the products of a reaction and the total enthalpy (heat content) of the reactants in that reaction.

Looking at the figure, we can see that the enthalpy of products is greater than the enthalpy of reactants, hence ∆Hrxn is positive as stated in the answer above.

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What is the functional group of a carboxylic acid?

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Answer:

A. Carbon double bonded to oxygen and a hydroxyl group (OH).

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Which of the following statements is TRUE?

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3 years ago

A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

rodikova [14]

Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = $1 g/cm^3=1 g/mL$

$1 mL= 1 cm^3$

$m=d\times v=1.0 g/mL\times 100.0 = 100.0 g$

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

$D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL$

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = $n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol$

Moles of water = $n_2=\frac{100.0 g}{18g/mol}=5.556 mol$

Mole fraction of phosphoric acid = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}$

$\chi_1=0.01803$

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}$

$\chi_2=0.9820$

$[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

$=\frac{0.1020 mol}{0.113 L}=0.903 M$

$[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}$

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

$=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg$

6 0

2 years ago

If a 500 ml sample of water contains 2.6 x 10-6 g of iron, what is the concentration of iron in ppb?

Anastasy [175]

The concentration of iron in ppb is 5.2 ppb

In atmospheric chemistry, PPM( parts per million) and PPB( parts per billion) units are used to express the concentration of gases. PPB stands for parts per billion, while PPM stands for parts of gas per million parts of air.

2.6 × 10⁻⁶ grams of iron in 500 ml of water.

Now, converting milliliter into liter:

500 mL = 500 mL × ( 1L / 1000 mL ) = 0.5 L

Now for ppm, mass must be in mg.

2.6 × 10⁻⁶ grams = 2.6 × 10⁻⁶ grams × ( 1000 mg / 1 grams ) = 2.6 × 10⁻³ mg

Then, the concentration of iron in ppm will be:

2.6 × 10⁻³ mg / 0.5 L = 5.2 × 10⁻³ ppm

Now, 1 ppm = 1000 ppb

Therefore,

5.2 × 10⁻³ ppm × 10³ = 5.2 ppb

Learn more about concentration here:

brainly.com/question/17206790#:~:text=15%20people%20helped-,Answer%3A,either%20solvent%20or%20total%20solution.

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