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kherson [118]
3 years ago
15

What is the volume of the solution produced when enough water is added to 42.0 g of mgcl2 ⋅ 6h2o to yield a solution that has a

cl- ion concentration of 0.500m?
Chemistry
1 answer:
DENIUS [597]3 years ago
5 0

Answer:

827 mL

Explanation:

To answer this question we use the <em>definition of Molarity</em>:

Molarity = mol / L

[Cl⁻] = mol Cl⁻ / L

Now we calculate the moles of Cl⁻ present in 42.0 g of MgCl₂⋅6H₂O:

Molar mass of MgCl₂⋅6H₂O = 24.3 + 2*35.45 + 6*18 = 203.2 g/mol

moles of Cl⁻ = 42.0 g MgCl₂⋅6H₂O ÷ 203.2 g/mol * \frac{2molCl^{-}}{1molMgCl_{2}.6H_{2}O} = 0.4134 mol Cl⁻

Finally we use the definition of molarity to calculate the volume:

0.500 M = 0.4134 mol Cl⁻ / xL

xL = 0.827 L =  827 mL

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Una masa de aire ocupa un volumen de5litro a una temperatura de 120c Cual será el nuevo volumen si la temperatura se reduce ala
zalisa [80]

Answer:

1. V_2=2.5L

2. V_2=8000mL

3. V_2=176.3L

Explanation:

¡Hola!

En este caso, dada la información para estos problemas, procedemos de la siguiente manera, basado en las leyes de los gases ideales:

1. Una masa de aire ocupa un volumen de 5 litros a una temperatura de 120 °C Cual será el nuevo volumen si la temperatura se reduce a la mitad:

Aqui, utilizamos la ley de Charles, asegurándonos que la temperatura está en Kelvin:

\frac{T_2}{V_2} =\frac{T_1}{V_1} \\\\V_2 =\frac{V_1T_2}{T_1} \\\\V_2 =\frac{5L*196.5K}{393K} \\\\V_2=2.5L

2. Un gas ideal ocupa un volumen de 4000 ml a una presión absoluta de 1500 kilo pascal Cual será la presión si el gas es comprimido lentamente hasta 750 kilo pascal a temperatura constante?

Aquí, utilizamos la ley de Boyle, dado que la temperatura se mantiene constante, calculando el volumen, ya que lo que se da es la presión final:

\neq P_2V_2=P_1V_1\\\\V_2=\frac{P_1V_1}{P_2}\\\\ V_2=\frac{4000mL*1500kPa}{750kPa}\\\\V_2=8000mL

3. Un gas ocupa un volumen de 200 litros a 95°C y 782 mmHg Cual será el volumen ocupado por dicho gas a 65°C y 815 mmHg:

Aquí, utilizamos la ley combinada de los gases ideales, asegurándonos que las temperaturas están en Kelvin:

\frac{T_2}{P_2V_2} =\frac{T_1}{P_1V_1} \\\\V_2 =\frac{P_1V_1T_2}{P_2T_1} \\\\V_2 =\frac{782mmHg*200L*338K}{815mmHg*368K}\\\\V_2=176.3L

¡Saludos!

6 0
3 years ago
When there are 0.0814 moles of hockey pucks how many hockey pucks are there?
Dovator [93]

Answer:

4.9 x 10²²hockey pucks

Explanation:

Given parameters:

Number of moles of hockey  = 0.0814moles

Unknown:

Number of pucks there = ?

Solution:

A mole of a substance is made up of Avogadro's number of particles.

Therefore;

       1 mole of hockey pucks will contain 6.02 x 10²³ hockey pucks

       0.0814 mole of hockey pucks will contain :

                              0.0814 x 6.02 x 10²³ = 4.9 x 10²²hockey pucks

8 0
3 years ago
Find the number of moles of water that can be formed if you have 182 mol of hydrogen gas and 86 mol of oxygen gas.
frutty [35]
Water has a chemical formula of H2O. This means that for every 2 moles of hydrogen and 1 mole of oxygen, one mole of water will be formed.

Note that hydrogen gas and oxygen gas are both biatomic molecules. 

 (1)                   (182 mol H2) x (1 mol H2O/ 1 mol H2) = 182 mol H2O
 (2)                   (86 mol O2) x (2 mol H2O / 1 mol O2) = 172 mol H2O

We choose the smaller number of the two as the answer to this item. Thus, the answer to this question is 172 mol of H2O can be formed out of the given quantities. 
3 0
3 years ago
If you start with 100 grams of hydrogen-3, how many grams will you have after 24.6 years?
svet-max [94.6K]

Answer:

The mass left after 24.6 years is 25.0563 grams

Explanation:

The given parameters are;

The mass of the hydrogen-3 = 100 grams

The half life of hydrogen-3 which is also known as = 12.32 years

The formula for calculating half-life is given as follows;

N(t) = N_0 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{\frac{1}{2} }} }

Where;

N(t) = The mass left after t years

N₀ =  The initial mass of the hydrogen-3 = 100 g

t = Time duration of the decay = 24.6 years

t_{\frac{1}{2} } = Half-life = 12.32 years

N(24.6) = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{24.6}{12.32}} } = 25.0563

The mass left after 24.6 years = 25.0563 grams.

4 0
3 years ago
The gas carbon dioxide is a pure substance. Which of the following is true about carbon dioxide? (5 points)
iogann1982 [59]
The proportion of carbon and oxygen is different in samples of the gas
5 0
4 years ago
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