I really need help with this assignment

Chemistry

1 answer:

Mumz [18]3 years ago

8 0

Answer:

A. Sublimation of camphor

You might be interested in

The bonds in the compound MgSO4 can be described as

Butoxors [25]

C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond

7 0

3 years ago

How do the chemical characteristics of carbon affect the characteristics of organic molecules?

Tom [10]

Answer:

The chemical characteristics of carbon affect the characteristics of organic molecules due to its tetravalent nature. It has four valence electrons in which it shares with other elements in order to form an octet configuration.

Carbon atoms are also capable of forming double and triple bonds with other atoms. These properties help determine the functional group present and gives us a knowledge of the chemical features such as polarity, melting and boiling present in the compound.

4 0

3 years ago

Anions have more ___ then __

never [62]

C protons, elctrons 093939393

6 0

3 years ago

H2SO4 + 2NaNO2 → 2HNO2 + Na2SO4

dem82 [27]

<h3>Answer:</h3>

23.459 g NaNO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄

[Given] 24.14714 g Na₂SO₄

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂

Molar Mass of Na - 22.99 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol

Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 24.14714 \ g \ Na_2SO_4(\frac{1 \ mol \ Na_2SO_4}{142.05 \ g \ Na_2SO_4})(\frac{2 \ mol \ NaNO_2}{1 \ mol \ Na_2SO_4})(\frac{69.00 \ g \ NaNO_2}{1 \ mol \ NaNO_2})$
Multiply/Divide: $\displaystyle 23.4587 \ g \ NaNO_2$