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4 years ago

7

The concentration of H30 in your swimming pool is 4 x 10^-5 M. What is the pH of your pool? O a.-4.4 Ob.4.4 Oc. 0.00004 O d.-0.6

O e.0.6

1 answer:

mariarad [96]4 years ago

8 0

Answer:

b. 4.4

Explanation:

pH is related to the concentration of H₃O⁺ through the following equation:

pH = -log([H₃O⁺]) = -log(4 x 10⁻⁵)

pH = 4.4

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What is the volume of 9.5 g fluorine gas, F2, at STP?

Tju [1.3M]

Answer:

5.6L

Explanation:

At STP, the pressure and temperature of an ideal gas is

P = 1 atm

T = 273.15k

Volume =?

Mass = 9.5g

From ideal gas equation,

PV = nRT

P = pressure

V = volume

n = number of moles

R = ideal gas constant =0.082J/mol.K

T = temperature of the ideal gas

Number of moles = mass / molar mass

Molar mass of F2 = 37.99g/mol

Number of moles = mass / molar mass

Number of moles = 9.5 / 37.99

Number of moles = 0.25moles

PV = nRT

V = nRT/ P

V = (0.25 × 0.082 × 273.15) / 1

V = 5.599L = 5.6L

The volume of the gas is 5.6L

5 0

3 years ago

The substances are placed in separate containers at room temperature, and each container is gradually cooled. Which of these sub

algol13

Answer:

Benzene

Explanation:

it's freezing point is 5.5 C which is greater than 0 C

3 0

3 years ago

A piece of plastic has a mass of 15 g and volume of 5 cm^3. what is the density........... g/cm^3

UNO [17]

Answer:

3g/cm cube

Explanation:

The most accurate way to calculate the density of any solid, liquid or gas is to divide its mass in kilograms by its volume (length × width × height) in cubic metres. The unit for density is kg/m 3. The density of water is approximately 1000 kg/m 3 and the density of air is approximately 1.2 kg/m 3.

15 divide by 5 = 3

4 0

3 years ago

Freeze-drying is a process used to preserve food. If strawberries are to be freeze-dried, then they would be frozen to -80.00 °C

katrin2010 [14]

Answer:

1. 389 kJ; 2. 7.5 µg; 3. 6.25 days

Explanation:

1. Energy required

The water is converted directly from a solid to a gas (sublimation).

They don't give us the enthalpy of sublimation, but

$\Delta_{\text{sub}}H = \Delta_{\text{fus}}H + \Delta_{\text{vap}}H = 6.01 + 40.68 = 46.69 \text{ kJ}\cdot\text{mol}^{-1}$

The equation for the process is then

Mᵣ: 18.02

46.69 kJ + H₂O(s) ⟶ H₂O(g)

m/g: 150

(a) Moles of water

$\text{Moles} = \text{150 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{8.324 mol}$

(b) Heat removed

46.69 kJ will remove 1 mol of ice.

$\text{Heat removed} = \text{8.234 mol} \times \dfrac{\text{46.69 kJ}}{\text{1 mol}} = \textbf{389 kJ}\\\text{It takes $\large \boxed{\textbf{389 kJ}}$ to remove 150 g of ice}$

2. Mass of water vapour in the freezer

For this calculation, we can use the Ideal Gas Law — pV = nRT

(a) Moles of water

Data:

$p = 1.00 \times 10^{-3}\text{ torr } \times \dfrac{\text{1 atm}}{\text{760 torr}} = 1.316 \times 10^{-6}\text{ atm}$

V = 5 L

T = (-80 + 273.15) K = 193.15 K

Calculation:

$\begin{array}{rcl}pV & = & nRT\\1.316 \times 10^{-6}\text{ atm} \times \text{5 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{193.15 K }\\6.6 \times 10^{-6} & = & 15.85n\text{ mol}^{-1} \\n & = & \dfrac{6.6 \times 10^{-6}}{15.85\text{ mol}^{-1}}\\\\& = & 4.2 \times 10^{-7} \text{ mol}\\\end{array}$

(b) Mass of water

$\text{Mass} = 4.2 \times 10^{-7} \text{ mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = 7.5 \times 10^{-6}\text{ g} = 7.5 \, \mu \text{g}\\\\\text{At any given time, there are $\large \boxed{\textbf{7.5 $\mu$g}}$ of water vapour in the freezer.}$

3. Time for removal

You must remove 150 mL of water.

It takes 1 h to remove 1 mL of water.

$\text{Time} = \text{150 mL} \times \dfrac{\text{1 h}}{\text{1 mL}} = \text{150 h} = \text{6.25 days}$

5 0

4 years ago

Which type of offspring are most likely the result of sexual reproduction?

Tresset [83]

Answer:

C

Explanation:

C.OD diverse offspring with the same genes as the parent

6 0

3 years ago