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Len [333]
3 years ago
8

2.00 liters of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated until a volume of 20.0 liters and a pressure

of 3.50 atmospheres is reached. What is the new temperature?
Chemistry
1 answer:
dedylja [7]3 years ago
6 0

Answer:  The new temperature is 10643 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 750.0 mm Hg = 0.98 atm   (760mmHg=1atm)

P_2 = final pressure of gas = 3.50 atm

V_1 = initial volume of gas = 2.00 L

V_2 = final volume of gas = 20.0 L

T_1 = initial temperature of gas = 25.0^oC=273+25.0=298.0K

T_2 = final temperature of gas = ?

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

\frac{0.98\times 2.00}{298.0K}=\frac{3.50\times 20.0}{T_2}

T_2=10643K

Thus the new temperature is 10643 K

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Answer:

              B. Green solution density is 1.06 g/ml and blue solution density is 1.20 g/ml

Explanation:

Density is given as,

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Red Solution,

                               D = 25 g / 25 mL

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                               D = 26.5 g / 25 mL

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Yellow Solution,

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Blue Solution,

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3 years ago
How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?
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<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<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
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  • Left to Right

<u>Chemistry</u>

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<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

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

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
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<u>Step 4: Check</u>

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

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

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