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weeeeeb [17]
3 years ago
11

A chemist determines by measurements that 0.0550 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of

nitrogen gas that participates.
Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
shtirl [24]3 years ago
6 0

The mass of nitrogen gas that participated in the chemical reaction is 1.54g

HOW TO CALCULATE MASS OF AN ELEMENT:

  • Mass of a substance can be calculated by multiplying the number of moles in mol of the substance by its molecular mass in g/mol. That is;

  • mass (M) = molar mass (MM) × number of moles (n)

According to this question, a chemist determines by measurements that 0.0550 moles of nitrogen gas (N2) participate in a chemical reaction.

  • The molecular mass of nitrogen gas (N2) = 14.01(2)

= 28.02g/mol

Hence, the mass of the nitrogen gas that participated in the chemical reaction is calculated as follows:

  • Mass (g) = 0.0550 mol × 28.02 g/mol

  • Mass = 1.5411

Therefore, the mass of nitrogen gas that participated in the chemical reaction is 1.54g

Learn more: brainly.com/question/18269198

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What is the empirical formula of C6H18O3?
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Answer:

The answer to your question is C₂H₆O

Explanation:

Data

Molecular formula = C₆H₁₈O₃

Empirical formula = ?

Empirical formula is defined as the simplest ratio of the elements that form part of a molecule.

Process

To find the empirical formula find the greatest common factor of the subscripts.

                             6    18   3   2

                             3      9  3   3

                              1     3    1   3

                                     1

The GCF is 3, so factor 3 of the molecular formula

                         3 ( C₂H₆O)  

The result is the empirical formula C₂H₆O

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3 years ago
What happens when there is an element and there is another element what is that called?
Oksanka [162]

It is called Chemical compound

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\large\rightarrow\pink{HCL}

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5 0
2 years ago
What is the color of the flame when magnesium burns in air ​
vfiekz [6]

Answer:

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white

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6 0
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In the galvanic cell Al(s) ǀ Al3+(aq, 1 M) ǀǀ Cu2+(aq, 1 M) ǀ Cu(s) which of the following changes will increase the cell potent
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The cathode electrode is copper. This means that the concentration of copper must be increased to increase the cell potential. This can be done by diluting the Al3+ solution. Increasing the surface area will not change the current but will increase the voltage. The answer is (D) I and III only.
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3 years ago
Read 2 more answers
What is the molar solubility of MgF2 in a 0.36 M Mg(NO3)2 solution? For MgF2, Ksp = 8.4 × 10^–8
Valentin [98]

Answer:

2.4 × 10⁻⁴ M

Explanation:

Step 1: Calculate the concentration of Mg²⁺ coming from Mg(NO₃)₂

Mg(NO₃)₂ is a strong electrolyte and the molar ratio of Mg(NO₃)₂ to Mg²⁺ is 1:1. The initial molar concentration of Mg²⁺ is 1/1 × 0.36 M = 0.36 M.

Step 2: Make an ICE chart for the solution of MgF₂

        MgF₂(s) ⇄ Mg²⁺(aq) + 2 F⁻(aq)

I                           0.36             0

C                           +S             +2S

E                         0.36+S         2S

The solubility product constant is:

Ksp = [Mg²⁺] × [F⁻]² = (0.36+S) × (2S)²

Since S <<< 0.36, 0.36+S ≈ 0.36.

Ksp = 0.36 × 4S² = 8.4 × 10⁻⁸

S = 2.4 × 10⁻⁴ M

7 0
3 years ago
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