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julsineya [31]
1 year ago
7

if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d

ata.
Chemistry
1 answer:
Pavlova-9 [17]1 year ago
3 0

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g

We can calculate the mass percent of O in MgO using the following expression.

\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\%  = 40.0 \%

You can learn more about mass percent here: brainly.com/question/14990953

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Answer: The concentration of Cl^- ions in the resulting solution is 1.16 M.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of the CaCl_2

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of the AlCl_3

We are given:

n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL  

Putting all the values in above equation, we get

M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M

The concentration of Cl^- ions in the resulting solution will be same as the molarity of solution which is 1.16 M.

Hence, the concentration of Cl^- ions in the resulting solution is 1.16 M.

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3 years ago
Which chemical family tends to get rid of electrons by forming a compound
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2 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>

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<u>Stoichiometry</u>

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<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

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