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mihalych1998 [28]
3 years ago
5

Calculate mass of oxygen

Chemistry
1 answer:
nignag [31]3 years ago
3 0
15.999 is the mass of oxygen
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When the white part is on the left the moon is.??​
Kazeer [188]

Answer:

Last Quarter also called Third Quarter.

Explanation:

6 0
2 years ago
How many molecules are in 85g of silver nitrate?
maksim [4K]
<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<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
  • Writing Compounds
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

85 g AgNO₃ (silver nitrate)

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

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})
  2. Multiply/Divide:                                                                                                \displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

6 0
2 years ago
I want one of the cute girls is this all to talk to me what do I have to do
Ivanshal [37]
Talk to them back don't be scared
4 0
3 years ago
Read 2 more answers
A sample of gas initially has a volume of 2.25 L at 350 K and a pressure of 1.75 atm. What will be sample pressure if the volume
IRINA_888 [86]

Answer:

8.44 atm

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 2.25 L

Initial temperature (T₁) = 350 K

Initial pressure (P₁) = 1.75 atm

Final volume (V₂) = 1 L

Final temperature (T₂) = 750 K

Final pressure (P₂) =?

The final pressure of the gas can be obtained as illustrated below:

P₁V₁/T₁ = P₂V₂/T₂

1.75 × 2.25 / 350 = P₂ × 1 / 750

3.9375 / 350 = P₂ / 750

Cross multiply

350 × P₂ = 3.9375 × 750

350 × P₂ = 2953.125

Divide both side by 350

P₂ = 2953.125 / 350

P₂ = 8.44 atm

Thus, the final pressure of the gas is 8.44 atm.

7 0
2 years ago
A mixture was found to contain 1.05 g if silicone dioxide, 0.69 g cellulose, 1.82 g of calium carbonate. what percent of calcium
djverab [1.8K]
1,05+ 0,69 + 1,82 = 3,56g of mixture

%CaCO₃: 1,82/3,56×100% = 0,5112×100% = 51,12%
5 0
2 years ago
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