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3 years ago

Which shows the correct relationship between the molecular formula and the empirical formula for C2H10N2?

Chemistry

2 answers:

sergij07 [2.7K]3 years ago

6 0

Answer:

Explanation:

SIZIF [17.4K]3 years ago

3 0

The correct answer is A. molecular formula = 2 x empirical formula

The given compound C2H10N2 is the molecular formula because the ratios of the elements are not reduced into lowest terms (empirical formula). To get the empirical formula, the ratios must be reduced to the smallest possible integral ratio which gives CH5N. As a result, the ratio between both formulas is 2.

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The reaction for the combustion of methane is shown below. If 28.70g of methane

antiseptic1488 [7]

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2 years ago

When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu

IgorLugansk [536]

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3 years ago

A block of iron has a mass of 150 g and a volume of 20 ml. What is the density of iron? *

shtirl [24]

Answer:

<h3>The answer is 7.50 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass of iron = 150 g

volume = 20 mL

We have

$density = \frac{150}{20} = \frac{15}{2} \\$

We have the final answer as

Hope this helps you

4 0

2 years ago

Read 2 more answers

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vagabundo [1.1K]

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3 years ago

PLEASE HELP ASAP!

choli [55]

Answer:

$\large \boxed{\text{D. 710 g}}$

Explanation:

1. Calculate the molar mass of Na₂SO₄

$\begin{array}{ccc}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 23 & 46\\\text{1S} & 32 & 32\\\text{4O}&16 & 64\\&\text{TOTAL =} & \mathbf{142}\\\end{array}$

The molar mass of Na₂SO₄ is 142 g/mol.

2. Calculate the moles of Na₂SO₄

$\text{Moles of Na$_{2}$SO}_{4} = \text{2.5 L solution} \times \dfrac{\text{2.0 mol Na$_{2}$SO}_{4}}{\text{1 L solution}} = \text{5.0 mol Na$_{2}$SO}_{4}$

3. Calculate the mass of Na₂SO₄

$\text{Mass of Na$_{2}$SO}_{4} = \text{5.0 mol Na$_{2}$SO}_{4} \times \dfrac{\text{142 g Na$_{2}$SO}_{4}}{\text{1 mol Na$_{2}$SO}_{4}} = \text{710 g Na$_{2}$SO}_{4}\\\\\text{You need } \large \boxed{\textbf{710 g}} \text{ of Na$_{2}$SO}_{4}$

6 0

3 years ago