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dmitriy555 [2]
3 years ago
14

If a gas is cooled from 425.0K to 275.15K and the volume is kept constant what final pressure would result if the original press

ure was 770.0 mm Hg?
Chemistry
1 answer:
mr Goodwill [35]3 years ago
4 0

Answer: 498.51mmHg

Explanation:

First let us analyse what was given from the question:

T1 = 425K

T2 = 275.15K

P1 = 770 mmHg

P2 =?

Using the general gas equation

P1V1/T1 = P2V2/T2

But we were told from the question that the volume is constant. So, our equation becomes:

P1/T1 = P2/T2

770/425 = P2 /275.15

Cross multiply to express in linear form:

P2 x 425 = 770 x 275.15

Divide both side by 425, we have:

P2 = (770 x 275.15) /425

P2 = 498.51mmHg

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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
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

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<h3>Explanation:</h3>

<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>

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Where are the metalloids located on the periodic table?
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Answer: The metalloids are a group of elements in the periodic table. They are located to the right of the post-transition metals and to the left of the non-metals. Metalloids have some properties in common with metals and some in common with non-metals.

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Now, replace one hydrogen on terminal carbon with -OH group and convert it into Butanol.

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With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?
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\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

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Thus CuCl_2 is the limiting reagent as it limits the formation of product and Zn is the excess reagent.

As 1 mole of CuCl_2 give = 1 mole of ZnCl_2

Thus 0.052 moles of CuCl_2 give =\frac{1}{1}\times 0.052=0.052moles  of ZnCl_2

Mass of ZnCl_2=moles\times {\text {Molar mass}}=0.052moles\times 136g/mol=7.07g

Thus 7.07 g of ZnCl_2 will be produced from the given masses of both reactants.

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