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2 years ago

Predict the products of the double replacement reactions given. Check to see that the equations are

Chemistry

1 answer:

Angelina_Jolie [31]2 years ago

4 0

Answer:

Option (2)

Explanation:

The two compounds formed will be AgCl and NaNO₃.

We can see that this will result in a balanced equation, so the answer is Option (2).

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Which state of matter does this image represent? Image of water Solid Liquid Gas Plasma

Nikitich [7]

The state of matter is liquid

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3 years ago

Consider the nuclear equation below.

Dominik [7]

²³⁵₉₂U → ²³¹₉₀Th + ⁴₂He

D) ²³¹₉₀Th

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3 years ago

Read 2 more answers

O The ester functional group is symbolized as || R-C-R True False

nevsk [136]

Answer:

False

The ester functional group is not symbolized as R-C-R

Explanation:

The general formula of ester functional group is RCOOR′ where R represents a hydrogen atom or aryl group or an alkyl group while R' represents the alkyl group or an aryl group

Hence, the given statement is false

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

What volume of solution must be added to 4.0 mol of NaCl to make a 1.2 M solution?

Aleksandr [31]

Answer:

$\boxed {\boxed {\sf 3.3 \ liters}}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity=\frac{moles \ of \ solute}{liters \ of \ solution}}$

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.

molarity= 1.2 mol/L
moles of solute= 4.0 mol
liters of solution =x

Substitute the values into the formula.

$1.2 \ mol/L = \frac{4.0 \ mol}{x}$

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.

$\frac {1.2 \ mol/L}{1}=\frac{ 4.0 \ mol}{x}$