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3 years ago

The capability of a substance to be stretched into thin wires without breaking is known as —

Chemistry

2 answers:

lisov135 [29]3 years ago

8 0

Answer:

The answer is D, ductility

Explanation:

The definition of ductile is "(of a metal) able to be drawn out into a thin wire."

PSYCHO15rus [73]3 years ago

3 0

The answer is ductility

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Question 1 of 25<br> Which sentence describes energy?

Slav-nsk [51]

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energy is the capability or ability to do work

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1 year ago

Cl2+2KI→2KCl + I2 is an example of which type of reaction?

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3 years ago

Why are Carbon-14 and carbon-12 considered to be isotopes?

zhannawk [14.2K]

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3 years ago

A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?

Luba_88 [7]

The empirical formula is XeO₃.

<u>Explanation:</u>

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/ 131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

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: $\displaystyle 23.4587 \ g \ NaNO_2$

<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

8 0

3 years ago

Read 2 more answers