Atoms of the same element can have different properties. True False

1. A solution is made by mixing 50mL of 2.0M K2HPO4 and 25mL of 2.0M KH2PO4. The solution is diluted to a final volume of 200mL.

Kazeer [188]

Answer:

The pH of the final solution is 7.15

Explanation:

50 mL of 2.0 M of $K_2HPO_4$ and 25 mL of 2.0 M of $KH_2PO_4$ were mixed to make a solution

Final volume of the solution after dilution = 200 mL

Final concentration of $K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M$

Final concentration of $KH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M$

We use Hasselbach- Henderson equation:

$pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85$

Substituting the values:

$pH = 6.85+ log \frac{0.5}{0.25}pH = 6.85+ log 2pH = 6.85+ 0.3 = 7.15$

Therfore, the pH of the final solution is 7.15

5 0

3 years ago

Gaseous indium dihydride is formed from the elements at elevated temperature:

3241004551 [841]

Answer:

For 1: The value of $Q_p$ for above reaction is 36.83

For 2: The value of $Q_p$ for above reaction is 36.83

For 3: The equilibrium partial pressure of Indium is 0.126 atm

For 4: The equilibrium partial pressure of hydrogen gas is 0.094 atm

For 5: The equilibrium partial pressure of Indium dihydrogen is 0.018 atm

Explanation:

We are given:

Partial pressure of Indium gas = 0.0650 atm

Partial pressure of hydrogen gas = 0.0330 atm

Partial pressure of Indium dihydride = 0.0790 atm

The given chemical equation follows:

$ln(g)+H_2(g)\rightleftharpoons InH_2(g)$

Initial: 0.065 0.033 0.079

At eqllm: 0.065-x 0.033-x 0.079+x

For 1:

The expression of $Q_p$ for above reaction follows:

$Q_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$Q_p=\frac{0.079}{0.065\times 0.033}=36.83$

Hence, the value of $Q_p$ for above reaction is 36.83

For 2:

We are given:

$K_p$ of the reaction = 1.48

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $Q_{p}>K_p$ for the given reaction, the reaction is reactant favored.

Hence, the reaction proceed in the backward direction to attain equilibrium

For 3:

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$1.48=\frac{(0.079+x)}{(0.065-x)\times (0.033-x)}\\\\x=-0.061,0.835$

Neglecting the value of x = 0.835 because the reaction is going backwards. So, by taking this value, the pressure of the reactants will decrease

So, equilibrium partial pressure of Indium = (0.065 - x) = [0.065 - (-0.061)] = 0.126 atm

For 4:

The equilibrium partial pressure of hydrogen gas = (0.033 - x) = [0.033 - (-0.061)] = 0.094 atm

For 5:

The equilibrium partial pressure of Indium dihydrogen = (0.079 + x) = [0.079 + (-0.061)] = 0.018 atm

6 0

3 years ago

The electrolysis of water has the equation 2H2O (l) → 2H2 (g) + O2 (g). What type of reaction is this?

Sonja [21]

Answer: The correct option is C) decomposition

Explanation:

Single displacement reaction is defined as the reaction where a more reactive element displaces a less reactive element from its salt solution.

$AX+B\rightarrow BX+A$

Element B is more reactive than element A

Double displacement reaction is defined as the reaction where an exchange of ions takes place in a solution

$AX+BY\rightarrow AY+BX$

Decomposition reaction is defined as the reaction where a single large chemical species breaks down into two or more smaller chemical species.

$AB\rightarrow A+B$

Synthesis reaction is defined as the reaction where two or more smaller chemical species combine together in their elemental state to form a single large chemical species.

$A+B\rightarrow AB$

For the given chemical reaction:

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$

Here, water is breaking down into two smaller species of oxygen and hydrogen gas. Thus, it is a decomposition reaction.

Hence, the correct option is C) decomposition

4 0

3 years ago

A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t

Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

$\frac{P1}{T1} =\frac{P2}{T2}$