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Alex73 [517]
2 years ago
14

How many moles of oxygen are in 2.00 moles of NaCIO4?​

Chemistry
1 answer:
Klio2033 [76]2 years ago
3 0

Answer: 8 Moles of Oxygen

Explanation: The formula shows that there are (exactly) 4 oxygen atoms in each mole or formula unit. Therefore 4 * 2.00 = 8.

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Friction is generated when _____ interfere with each other on sliding surfaces. solids molecules forces fluids
9966 [12]
Friction is generated when sliding
5 0
3 years ago
HELP PLEASE I NEED HELP THANKS I LOVE U
Pani-rosa [81]

Answer:

0.500-Molarity solution

Explanation:

4 0
3 years ago
What is the mass of 3.77mol of K3N?
AleksAgata [21]
<h3>Answer:</h3>

495 g K₃N

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.77 mol K₃N

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

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

  1. Set up:                       \displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})
  2. Multiply/Divide:         \displaystyle 495.039 \ g \ K_3N

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

5 0
3 years ago
"determine the mass of oxygen" in a 7.9 g sample of al2(so4)3.
jeyben [28]

Answer:

              4.43 g of Oxygen

Explanation:

As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;

                          2 Moles of Aluminium

                          3 Moles of Sulfur

                          12 Moles of Oxygen

Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,

          342.15 g ( 1 mole) of Al₂(SO₄)₃ contains  =  192 g (12 mole) of O

So,

                         7.9 g of Al₂(SO₄)₃ will contain  =  X g of O

Solving for X,

                       X  =  (7.9 g × 192 g) ÷ 342.15 g

                      X =  4.43 g of Oxygen

7 0
3 years ago
R indicates round seeds that are dominant over the wrinkled variety represented by R for each numbered individual in the pedigre
hichkok12 [17]

Answer:

Explanation:X and Y

4 0
2 years ago
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