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2 years ago

En 200 gramos de carburo de potasio ,cuya fórmula química es K4C , Cuántos moles de esa sustancia existen ?

Chemistry

1 answer:

WARRIOR [948]2 years ago

3 0

Answer:

i would happily help but i don't understand that language sorry

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Which of the following can’t be separated into simpler substances or converted into another substance by chemical process?

timofeeve [1]

Your answer would be an Element can't be separated into simpler substances or converted into another substance by chemical process.

Hope that helps!!!

8 0

3 years ago

Refer to the example about diatomic gases A and B in the text to do problems 19 - 27. How many grams in 3 moles of A2?

dedylja [7]

B.) 3

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(GAME OF THRONES RULES

7 0

3 years ago

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The equilibrium constant for the gas phase reaction N2 (g) + O2 (g) ⇌ 2NO (g) is Keq = 4.20 ⋅ 10-31 at 30 °C. At equilibrium, __

pychu [463]

Answer:

At equilibrium, reactants predominate.

Explanation:

For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:

$Keq = \frac{[NO]^{2}}{[O_{2} ][N_{2}]}$

Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as 4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.

5 0

3 years ago

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Compared to a 0.1 M aqueous solution of NaCl, a 0.8 M aqueous solution of NaCl has a

Fudgin [204]

The correct answer is option 2. A 0.8 M aqueous solution of NaCl has a higher boiling point and a lower freezing point than a 0.1 M aqueous solution of NaCl. This is explained by the colligative properties of solutions. For the two properties mentioned, the equation for the calculation of the depression and the elevation is expressed as: ΔT = -Km and <span>ΔT = Km, respectively. As we can see, concentration and the change in the property has a direct relationship.</span>

3 0

3 years ago

A chemist uses hot hydrogen gas to convert chromium(III) oxide to

Darya [45]

Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$

Explanation:

The reaction equation for given reaction is as follows.

$Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O$

Here, 1 mole of $Cr_{2}O_{3}$ reacts with 3 moles of $H_{2}$ .

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide $(Cr_{2}O_{3})$ is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of $Cr_{2}O_{3}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol$

Now, moles of $H_{2}$ .given by 0.5 mol of $Cr_{2}O_{3}$ is calculated as follows.

$0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}$

As molar mass of $H_{2}$ is 2.016 g/mol. Therefore, mass of $H_{2}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g$

Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$ .

7 0

3 years ago