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3 years ago

How many moles are in 55.0 g calcium nitrate?

Chemistry

1 answer:

Alex17521 [72]3 years ago

5 0

Answer: 0.335 mols Ca(NO3)2

Explanation: To go from grams to mols you divide grams by the molar mass of the substance so, 55/164.088 = 0.335

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Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.

attashe74 [19]

Answer:

6.23 KOH 90% son necesarios

Explanation:

Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.

Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:

Equivalentes KOH:

0.100L * (1eq / L) = 0.100eq = 0.100moles

Gramos KOH -Masa molar: 56.1056g/mol-:

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

KOH 90%:

5.61g KOH * (100g KOH 90% / 90g KOH) =

<h3>6.23 KOH 90% son necesarios</h3>

8 0

2 years ago

What is the color of litmus solution in sodium hydroxide?

Ronch [10]

Answer:

Sodium Hydroxide turns blue litmus red .

4 0

3 years ago

State of matter that keeps its shape an volume when it is placed in a different container

luda_lava [24]

Solids are the only ones that keep their shape and volume no matter the container.

3 0

3 years ago

Read 2 more answers

A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium,

Scorpion4ik [409]

Answer: The value of $K_c$ is coming out to be 0.412

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $Sb_2S_3$

Given mass of $Sb_2S_3$ = 1.00 kg = 1000 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $Sb_2S_3$ = 339.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol$

For hydrogen gas:

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol$

For hydrogen sulfide:

Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol$

The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:

$Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)$

Initial: 2.944 5

At eqllm: 2.944-x 5-3x 2x 3x

We are given:

Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

$\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712$

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

The expression of $K_c$ for above equation, we get:

$K_c=\frac{[H_2S]^3}{[H_2]^3}$

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

$K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412$

Hence, the value of $K_c$ is coming out to be 0.412

3 0

3 years ago

If the reaction of 9.87 moles of potassium with excess hydrobromic acid produced a 79.8% yield of hydrogen gas, what was the act

Fed [463]

Answer is: 3,94 of hydrogen gas.
Chemical reaction: 2K + 2HBr → 2KBr + H₂.
n(K) = 9,87 mol.
n(H₂) = ?.
from reaction: n(K) : n(H₂) = 2 : 1.
9,87 mol : n(H₂) = 2 : 1
n(H₂) = 4,935 mol for 100% yield of reaction
n(H₂) = 4,935 · 0,798 = 3,94 mol for 78,9 % yield of reaction.
n - amount of substance

7 0

3 years ago