What is the pH of a solution that has a hydrogen ion concentration of 1.0 * 10 -9 M? Show your calculation.

Chemistry

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

9.00

Explanation:

Data:

[H⁺] = 1.0 × 10⁻⁹ mol·L⁻¹

Calculation:

pH = -log[H⁺] = -log(1.0 × 10⁻⁹) = -log(1.0) - log(10⁻⁹) = -0.00 - (-9) = -0.00 + 9 = 9.00

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What are mixtures that have evenly mixed groups of particles that do not settle out called?

Alisiya [41]

Homogeneous mixtures.
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3 years ago

A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon.

zalisa [80]

Answer:

a) Same

b) Nitrogen

c) Same

d) Nitrogen

Explanation:

The formula for partial pressure of a gas is equal to

$p_B = n_B \frac{RT}{V}$

Here nB is the number of moles .

The number of moles for both the gases are same and hence the partial pressure for the two gases will also be same.

b) The greater average velocity is calculated by using following formula

$v_{RMS} = \sqrt{3RTM}$

Here M is the molar mass.

Molar mass of nitrogen is greater than the molar mass of xenon and hence nitrogen will have higher greater average velocity

c) As we know, the average kinetic energy of gas particles is dependent on the absolute temperature of gas and if all the gases are at same temperature, their kinetic energy will also be same. Since nitrogen and xenon are at same temperature, their kinetic energy will be same

d) Effusivity is depended directly on the thermal conductivity, density and and the specific heat capacity.

All these three parameters are higher in case of nitrogen. Thus, it will effuse first

4 0

3 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$