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2 years ago

1. How many MOLECULES are contained in 80.0 grams of sodium hydroxide?

Chemistry

1 answer:

Anna11 [10]2 years ago

6 0

Answer:

2 molecules

Explanation:

In 80 grams there will be 2 molecules of NaOH, so mass of Sodium would be 23 x 2 = 46g.

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PLEASE HELP DUE SOON

jonny [76]

Answer:

This is all true if the atom has to be neutral.

Also what does V mean?

Helium: one shell with 2 neutrons and 2 protons in the center, with 2 electrons in the first shell.

Lithium: two shells with 4 neutrons and 3 protons in the center, with 2 electrons in the first shell, and 1 electron in the second shell.

Nitrogen: two shells with 7 neutrons and 7 protons in the center, with 2 electrons in the first shell, and 5 electrons in the second shell.

Flourine: two shells with 9 protons and 10 neutrons in the center, with 2 electrons in the first shell, and 7 electrons in the second shell.

Neon: two shells with 10 neutrons and 10 protons in the center, with 2 electrons in the first shell, and 8 electrons in the second shell.

Boron: two shells with 6 neutrons and 5 protons in the center, with 2 electrons in the first shell, and 3 electrons in the second shell.

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2 years ago

What is called when a paragraph studies data carefully

Zigmanuir [339]

Paraphrasing and summarizing

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2 years ago

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25 POINTS:)

Digiron [165]

I suppose that the answer is A

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2 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: $\displaystyle 23.4587 \ g \ NaNO_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We need 5 sig figs (instructed).</em>

23.4587 g NaNO₂ ≈ 23.459 g NaNO₂

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2 years ago

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Blank Na2CO3 + blank Ag3P --> blank Na3P + blank Ag2CO3

mezya [45]

Answer: __3__Na2CO3 + __2__Ag3P → __2__Na3P + __3__Ag2CO3

Explanation:

In balancing equations you need to make sure the number of atoms before the reaction should be equal to the number of atoms after the reaction.

Before the reaction: We had 6Na, 9CO, 6Ag and 2p

After the reaction: We have 6Na, 9CO, 6Ag and 2p

And now we are sure that atoms before the reaction are equal to atoms after the reaction

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2 years ago