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3 years ago

What volume in L of a 0.724 M Nal solution contains 0.405 mol of Nal?

Chemistry

2 answers:

bekas [8.4K]3 years ago

6 0

Answer:

Explanation:

I added the pictures for anyone struggling with that aspect

Setler [38]3 years ago

4 0

Answer:

0.559 L

Explanation:

Step 1: Given data

Moles of sodium iodide (n): 0.405 mol

Molar concentration of sodium iodide (M): 0.724 M (0.724 mol/L)

Step 2: Calculate the volume of solution (V)

The molarity is equal to the moles of solute divided by the liters of solution.

M = n/V

V = n/M

V = 0.405 mol/(0.724 mol/L) = 0.559 L

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The effect of a catalyst on an equilibrium is to ________. Select one: a. increase the rate of the forward reaction only b. slow

Georgia [21]

Answer:

Explanation:

A catalyst is a substance which alters the rate of a chemical reaction. We check for the correctness of each of the options as follows:

A. Is wrong.

A catalyst can increase the rate of the forward and backward reaction

B. Is wrong

A catalyst does not slow the reverse reaction only. This particular case is the case of a negative catalyst

C is wrong

A catalyst has no effect on the equilibrium nor the equilibrium constant

D is wrong

Catalyst has no effect on equilibrium value

E is correct

Although Catalysts has no effect on equilibrium or its constant value, it can increase the the rate at which equilibrium is achieved by speeding up the reaction through bringing down the activation energy

5 0

3 years ago

En un vaso de precipitado de un litro se coloca exactamente 500 mL de agua destilada a temperatura ambiente y se realizan dos ex

Serga [27]

Answer:

1) La masa del agua a temperatura ambiente es de 500 gramos, 2) La masa del agua cuando se congela es de 500 gramos, 3) La masa de agua que queda después de la evaporación es de 400 gramos, 4) Se ha evaporado 100 gramos de agua.

Explanation:

1) ¿Cuál es la masa de agua a temperatura ambiente?

Podemos determinar la masa inicial del agua ( $m_{o}$ ), medido en gramos, al conocer su densidad ( $\rho_{w}$ ), medida en gramos por mililitro, y volumen inicial ocupado en el vaso de precipitado ( $V_{o}$ ), medido en mililitros, a partir de la siguiente expresión:

$m_{o} =\rho_{w}\cdot V_{o}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{o} = 500\,mL$ , entonces:

$m_{o} = \left(1\,\frac{g}{mL} \right)\cdot (500\,mL)$

$m_{o} = 500\,g$

La masa del agua a temperatura ambiente es de 500 gramos.

2) ¿Cuál es la masa de agua cuando se congela?

Puesto que el proceso de congelación no implica transferencia de masa, la masa de agua se conserva al transformarse en hielo. Por tanto, la masa resultante es de 500 gramos.

3) ¿Cuál es la masa de agua que queda después de la evaporación?

Durante la evaporación una parte del agua es transferida al aire, entonces podemos calcular la masa final ( $m_{f}$ ), medido en gramos, de la sustancia al multiplicar el volumen final ( $V_{f}$ ), medido en mililitros, por la densidad del agua ( $\rho_{w}$ ), medida en gramos por mililitro,. Es decir,

$m_{f} =\rho_{w}\cdot V_{f}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{f} = 400\,mL$ , entonces:

$m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)$

$m_{f} = 400\,g$

La masa de agua que queda después de la evaporación es de 400 gramos.

4) ¿Qué masa de agua se evaporó?

Determinamos que la masa evaporada de agua ( $m_{v}$ ), medida en gramos, es igual a la diferencia entre las masas inicial y final, ambas medidas en gramos:

$m_{v} =m_{o}-m_{f}$

Si $m_{o} = 500\,g$ y $m_{f} = 400\,g$ , entonces tenemos que:

$m_{v} = 500\,g -400\,g$

$m_{v} = 100\,g$

Se ha evaporado 100 gramos de agua.

5 0

3 years ago

The pressure of nitrogen gas at 35°C is changed from 0.89 atm to 4.3 atm. What will be its final temperature in Kelvin?

Alja [10]

Answer: The final temperature in Kelvin is 1488

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=0.89atm\\T_1=35^0C=(35+273)K=308K\\P_2=4.3atm\\T_2=?$

Putting values in above equation, we get:

$\frac{0.89}{308}=\frac{4.3}{T_2}\\\\T_2=1488K$

Hence, the final temperature in Kelvin is 1488

8 0

3 years ago

Nicotine, a component of tobacco, is composed of c, h, and n. a 4.200-mg sample of nicotine was combusted, producing 11.394 mg o

Rudiy27

Answer:

Empirical Formula = C₅H₇N₁

Solution:

Data Given:

Mass of Nicotine = 4.20 mg = 0.0042 g

Mass of CO₂ = 11.394 mg = 0.011394 g

Mass of H₂O = 3.266 mg = 0.003266 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.011394 ÷ 0.0042) × (12 ÷ 44) × 100

%C = (2.7128) × (12 ÷ 44) × 100

%C = 2.7128 × 0.2727 × 100

%C = 73.979 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.003266 ÷ 0.0042) × (2.02 ÷ 18.02) × 100

%H = (0.7776) × (2.02 ÷ 18.02) × 100

%H = 0.7776 × 0.1120 × 100

%H = 8.709 %

%N = 100% - (%C + %H)

%N = 100% - (73.979 % + 8.709%)

%N = 100% - 82.688%

%N = 17.312 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 73.979 ÷ 12.01

Moles of C = 6.1597 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 8.709 ÷ 1.01

Moles of H = 8.6227 mol

Moles of N = %N ÷ At.Mass of O

Moles of N = 17.312 ÷ 14.01

Moles of N = 1.2356 mol

Step 3: Find out mole ratio and simplify it;

C H N

6.1597 8.6227 1.2356

6.1597/1.2356 8.6227/1.2356 1.2356/1.2356

4.985 6.978 1

≈ 5 ≈ 7 1

Result:

Empirical Formula = C₅H₇N₁

6 0

3 years ago

Complete the reaction (NH4)2SO4(aq)+BaCl2(aq)

rosijanka [135]

(NH4)2SO4(aq) + BaCl2(aq) = BaSO4(s) + 2 NH4Cl(aq)
Reaction type: double replacement

5 0

4 years ago