A substance that does not dissolve in a solvent is said to be:

A 5.0 L sample of gas at 300. K is heated to 600. K. What will the new volume of the gas be?

Ainat [17]

Answer:

$V_2=10L$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required new volume by using the Charles' law as a directly proportional relationship between temperature and volume:

$\frac{V_2}{T_2} =\frac{V_1}{T_1}$

In such a way, we solve for V2 and plug in V1, T1 and T2 to obtain:

$V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{5.0L*600K}{300K}\\\\V_2=10L$

Regards!

4 0

3 years ago

Which is the smallest particle into which water (H2O) in a glass can be broken down and still remain water? A. a water molecule

Gnesinka [82]

Answer:-

A. A water molecule

Explanation:-

A molecule is the smallest particle of a compound that retains all it's chemical properties.

Here H2O is a compound. So the smallest particle that will retain all the chemical properties and still remain water is water molecule.

Atoms Hydrogen and Oxygen both have different chemical properties from water H2O and are thus different.

Hydrogen peroxide is different molecule from water. So it is also not water.

6 0

3 years ago

Read 2 more answers

Match the following phases of stellar evolution to its characteristic.

Artist 52 [7]

Answer:

im the only answer your gonna get

Explanation:

7 0

2 years ago

The figure shows a nutrition label. How much energy does this food contain?

RoseWind [281]

Answer:

i need help with that too.

Explanation:

3 0

3 years ago

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$

$K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}$

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

$Q= \dfrac{1}{Pso_2Po_2^{1/2}}$

Since we are dealing with liquids;

$Q= \dfrac{1}{1 \times 1^{1/2}}$

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0

3 years ago