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3 years ago

I need help with these two

Chemistry

1 answer:

bearhunter [10]3 years ago

3 0

Answer:

6. d, 7. a

Explanation:

6. Molarity is a number of moles solute in 1 L solution.

7. 1 L solution - 2.5 mol K2CO3

20 L - x mol K2CO3

x =20*2.5/1 = 50 mol K2CO3

Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol

99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.

Actually KCO3 does not exist, in reality it should be K2CO3.

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Calculate the pH of the resulting solution if 32.0 mL of 0.320 M HCl(aq) is added to (a) 42.0 mL of 0.320 M NaOH(aq). (b) 22.0 m

monitta

Answer:

a) pH = 11.5

b) pH= 3

Explanation:

a) lets calculate the number of moles of each reactant

moles of HCl = 32/1000 * 0.32 = 0.01024 mole

moles of NaOH = 42/1000 * 0.32 = 0.01344 moles

1 moles of HCl reacts with 1 moles of NaOH , so 0.01024 mole of HCl should react with 0.01024 moles of NaOH , but there is some excess NaOH.

excess NaOH= 0.01344 -0.01024 = 0.0032 moles

[H+]= $\frac{10^{-14} }{0.0032} = 3.125 *10^{-12}$

pH= -log [3.125*10^-12) = 11.5

b) moles of NaOH = 0.00924

excess HCl present = 0.01024 - 0.00924 =.001

so excess [H+] = 0.001

pH= -log( 0.001) = 3

3 0

3 years ago

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}$

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142$

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen = $d_{O_2}=\frac{1 m}{t}$

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane= $d_{CH_4}=\frac{y }{t}$

$\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142$

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0

3 years ago

The formula of Benadryl is C17H21NO. How many moles of C are present in 0.733 mol of Benadryl?

IrinaK [193]

Answer:

The formula of benadryl is C17H21NO. How many moles of C are present in 0.759 mole of benadryl;;. tom. Nov 27, 2015. 0.759 mols ben x (17 C atoms/1 mol ben) = ?;. DrBob222.

3 0

3 years ago

A 20 kg curling stone is sliding is sliding in a positive direction at 4 m/s. A second curling stone is sliding at the same spee

Colt1911 [192]

Answer:

The kinetic energy of the two curling stones is 320 J

Explanation:

Given;

mass of the first curling stone, m₁ = 20 kg

velocity of the first curling stone, v₁ = 4 m/s

velocity of the second curling stone, v₂ = - 4m/s

assuming the second curling stone to have equal mass with the first stone = 20 kg

The kinetic energy of the first curling stone is given by;

K.E₁ = ¹/₂m₁v₁²

K.E₁ = ¹/₂(20)(4)²

K.E₁ = 160 J

The kinetic energy of the second stone is given by;

K.E₂ = ¹/₂m₂v₂²

K.E₂ = ¹/₂ (20) (-4)²

K.E₂ = 160 J

Thus, the kinetic energy of the two curling stones is given by;

K.E = K.E₁ + K.E₂

K.E = 160 J + 160 J

K.E = 320 J

Therefore, the kinetic energy of the two curling stones is 320 J

5 0

3 years ago

An atom that loses an electron is called a?

Reptile [31]

It’s called an ion . An atom that loses electron is called an ion

3 0

3 years ago