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3 years ago

What type of reaction is represented by the following equation:

Chemistry

1 answer:

max2010maxim [7]3 years ago

6 0

Answer:

Acid-base

Explanation:

KOH is a strong base, and NH₄Cl is an acid. They react to form a salt and water.

KOH + NH₄Cl ⟶ NH₃ + KCl + H₂O

base acid salt water

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Which one is it? Please help

DedPeter [7]

Answer:

D: 60 grams.

Explanation:

If you look at the x-axis and where it says 50 degrees Celsius, you can see that it takes 20 grams of solute to be saturated at 100 GRAMS of solvent. However, the question is asking for the amount of solute to saturate a 300 GRAM solution.

This means you will have to multiply the amount of solute needed at 100 grams by 3 to get you to the amount needed for 300 grams.

20 times 3 will get you to your answer, 60 grams.

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4 years ago

What are other ways you can use wind energy to create electricity?

Phantasy [73]

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3 years ago

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Anna35 [415]

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3 years ago

What is the density of a bar of soap with a mass of 715g, and a volume of 866cm3?

mihalych1998 [28]

Answer:

0.826 g/cm³

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4 0

3 years ago

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

const2013 [10]

Answer:

$\boxed{\text{47.4 g}}$

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 17.03 32.00 18.02

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g: 70.1 70.1

Step 1. Calculate the moles of each reactant

$\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}$

Step 2. Identify the limiting reactant

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

$\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}$

From O₂:

The molar ratio of H₂O:O₂ is 6:5.

$\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}$

O₂ is the limiting reactant because it gives the smaller amount of H₂O.

Step 3. Calculate the theoretical yield.

$\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}$

6 0

3 years ago