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MAXImum [283]
4 years ago
9

You wash dishes for a chemistry laboratory to make extra money for laundry. You earn 12 dollars/hour, and each shift lasts 75 mi

nutes. Your laundry requires 12 quarters/load. How many loads of laundry will each shift pay for if the cost per load rises to 16 quarters?
Chemistry
2 answers:
natita [175]4 years ago
8 0

Each shift worked can pay for 3.75 loads of laundry if each load costs 16 quarters.

<h3>Further Explanation</h3>

This problem can be solved using simple dimensional analysis. The steps are:

  1. Sort the given. Identify the conversion factors or equalities that may be used. Identify what is required.
  2. Set up the dimensional analysis ensuring that units cancel out until only the desired unit is left. The equalities in the problem may be used as conversion factors.

STEP 1: Sort first the given in the problem to identify possible conversion factors to be used in the dimensional analysis.

The following equalities are given in the problem:

  • 1 shift = 75 minutes
  • 12 dollars = 1 hour
  • 1 load = 16 quarters
  • 1 dollar = 4 quarters

STEP 2: From these equalities, the following dimensional analysis can be set up:

no. \ of \ load \ = 1 \ shift \times \frac{75 \ min}{1 \ shift} \times {\frac{1 \ hr}{60 \ min} \times \frac{12 \ dollars}{1 \ hr} \times { \frac{4 \ quarters}{1 \ dollar} \times \frac{1 \ load}{16 \ quarters}

no. \ of \ load \ = 3.75 \ load

Therefore, one shift can pay for 3.75 loads of laundry.

<h3>LEARN MORE</h3>
  • Learn more about Unit Conversion brainly.com/question/1594497
  • Learn more about Dimensional Analysis brainly.com/question/8013893

Keywords: dimensional analysis, problem solving

PIT_PIT [208]4 years ago
4 0
You calculate the amount of loads of laundry as follows:

((6 x 0.25)/ load) x 10 loads = 15.00 total cost required for laundry 

<span>(6.00 / 60 min) x (75 min/shift) = 7.50 cost / shift </span>

15.00 / (7.50 / shift) = 2 loads of  laundry

Hope this answers the question.
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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Na2Co3 and BaCl2 form precipitate
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The three end stages of stars are Red Giant Phase, Second Red Giant Phase and White Dwarf Phase correct?
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3 years ago
How many grams of Al(no2)3 should be added to 1.3L of water to prepare a 2.0M solution
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Answer: 429 grams of Al(NO_2)_3  should be added to 1.3 L of water to prepare 2.0 M solution

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute Al(NO_2)_3 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{165g/mol} 

V_s = volume of solution in Liters  

Now put all the given values in the formula of molarity, we get

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3 years ago
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Misha Larkins [42]

Explanation:

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Answer:

Thus, 67.2 L of CO2 is formed at STP.

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