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Shkiper50 [21]
4 years ago
5

A sample of gas occupies 280 mL when the pressure is 560.00 mm Hg . If the temperature remains constant , what is the new pressu

re if the volume changes to 560.0 mL?
Chemistry
1 answer:
vichka [17]4 years ago
3 0

Answer : The new pressure if the volume changes to 560.0 mL is, 280 mmHg

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 560.00 mmHg

P_2 = final pressure = ?

V_1 = initial volume = 280 mL

V_2 = final volume = 560.0 mL

Now put all the given values in the above formula, we get:

560.00mmHg\times 280 mL=P_2\times 560.0 mL

P_2=280mmHg

Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg

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Answer:

The correct answer is B.

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one reaction that produces hydrogen gas can be represented by the unbalanced chemical equation Mg(s)+HCI(aq) -> MgCI(aq)+H2(g
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<h3>Answer:</h3>

128 g HCl

<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>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)

↓

[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)

[Given] 3.25 mol Mg

[Solve] x g HCl

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

[RxN] 2 mol Mg → 2 mol HCl

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol

<u>Step 3: Stoich</u>

  1. [S - DA] Set up:                                                                                                 \displaystyle 3.25 \ mol \ Mg(\frac{2 \ mol \ HCl}{2 \ mol \ Mg})(\frac{36.46 \ g \ HCl}{1 \ mol \ HCl})
  2. [S - DA] Multiply/Divide [Cancel out units]:                                                    \displaystyle 127.61 \ g \ HCl

<u>Step 4: Check</u>

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

127.61 g HCl ≈ 128 g HCl

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