What is the change in enthalpy for a reaction if the reactants have 270 kJ of energy and the products have 150 kJ of energy

A solution is prepared from 8 grams of acetic acid (CH2CO2H) and 725 grams of water. What is the molarity of this solution?

Andrews [41]

Hello!

First, we can assume the density of water to be 1 g/mL, so the volume of water would be 725 mL. The density of Acetic Acid (Pure) is 1,05 g/mL so 8 grams would represent 7,61 mL

Now we can apply the following conversion factor to calculate the molarity of the solution, using the molar mass of Acetic Acid:

$[CH_3CO_2H]=\frac{8 g_{CH_3CO_2H} }{725 mL + 7,61 mL}* \frac{1000 mL}{1 L}* \frac{1 mol_{CH_3CO_2H} }{60,05 g_{CH_3CO_2H} }$

$[CH_3CO_2H]=0,1818 M$

So, the concentration of acetic acid would be 0,1818 M

Have a nice day!

4 0

4 years ago

Read 2 more answers

1N2 + 3H2 -->

Hunter-Best [27]

Answer:

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

5 0

3 years ago

Prokaryotic cell division is a___________process involving onc chromosom.

Trava [24]

Answer:

It’s is a simple

Explanation:the prokaryotic cell involves a simple process which intern involves one chromosom

5 0

3 years ago

Two gas-containers, A and B, are connected with a valve. The first container, A, has a volume of 135 mL, and the second containe

stiv31 [10]

Answer:

85.5 mmHg is the pressure of the gas sample when the valve is opened.

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas in container A = 165 mmHg

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas in container A= $135 mL$

$V_2$ = final volume of gas = 135 mL + 117 mL = 252 mL

$T_1$ = initial temperature of gas in container A = $22.5^oC=273+22.5=295.5 K$

$T_2$ = final temperature of gas = $12.7^oC=273+12.7=285.7K$

Now put all the given values in the above equation, we get:

$P_2=\frac{P_1V_1\times T_2}{T_1\times V_2}$

$=\frac{165 mmHg\times 135 mL\times 285.7 K}{295.5 K\times 252 mL}$

$P_2=85.5 mmHg$

85.5 mmHg is the pressure of the gas sample when the valve is opened.

8 0

4 years ago

In a solution of H2SO4, rank the concentration of H2SO4, H+, SO42- from highest concentration to lowest concentration.

Softa [21]

Answer:

[H⁺] > [SO₄²⁻] > [H₂SO₄]

Explanation:

H₂SO₄ is a strong acid, which means that most of it ionizes in aqueous solution.

Since it is a diprotic acid (two hydrogen ions) its ionization occurs in two steps:

H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻ (aq) → H⁺(aq) + SO₄²⁻(aq)

Thus, almost all H₂SO₄ has ionized and its final concentration is almost nothing.

After the first ionization, the conentrations of H⁺(aq) and HSO₄⁻ are equal but by the second ionization more H⁺ ions are produced along with SO₄⁻.

You can show it as one step dissociation, assuming 100% dissociation (given this is a strong acid):

By the stequiometry you can build this table:

H₂SO₄ (aq) → 2H⁺(aq) + SO₄²⁻(aq)

Initial A 0 0

Change - x +2x +x

Equilibrium A - x 2x x

As explained, A - x is very low, and 2x is twice x. Thus,

The rank of the concentrations from highest to lowest is:

[H⁺] > [SO₄²⁻] > [H₂SO₄]

5 0

3 years ago