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Anastaziya [24]
2 years ago
12

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

uation P V = C , where C is a constant. Suppose that at a certain instant the volume is 500 cm 3 , the pressure is 200 kPa , and the pressure is increasing at a rate of 30 kPa/min . At what rate is the volume decreasing at this instant?
Chemistry
1 answer:
ad-work [718]2 years ago
5 0

Answer:

-75 cm^3/min

Explanation:

Given from Boyle's law;

PV=C

From product rule;

VdP/dt + PdV/dt = dC/dt

but dC/dt = 0, V= 500 cm^3, P= 200kPa, dP= 30kPa/min

PdV/dt = dC/dt - VdP/dt

dV/dt = dC/dt - VdP/dt/ P

substituting values;

dV/dt = 0 - (500 * 30)/200

dV/dt = -75 cm^3/min

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The graph shows the consumption of gasoline and diesel fuels by vehicles in the United States for a period of 40 years.
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B. It would lead to an increase of global temperatures.

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4. How many moles of gold (Au) are in a pure gold nugget having a mass of 25.0 grams.
ikadub [295]
<h3>Answer:</h3>

0.127 mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</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>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 25.0 g Au

[Solve] moles Au

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

[PT] Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. [DA] Multiply/Divide [Cancel out units:                                                          \displaystyle 0.126923 \ mol \ Au

<u>Step 4: Check</u>

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

0.126923 mol Au ≈ 0.127 mol Au

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2 years ago
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In an estuary, _____.
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B; Seawater mixes with freshwater so the water has intermediate salinity

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3 years ago
What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL
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Density is weight by volume.   

First.  If you divide the weight by density you can find the volume

Second you must convert the ML in to Liters.

\frac{321 \frac{g}{1}}{0.84 \frac{g}{mL}} = \frac{321(g)(mL)}{0.84g}=\frac{321mL}{0.84}=382.14mL

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\frac{1L}{1000mL}

(382.14mL)( \frac{1L}{1000mL})= \frac{(382.14mL)(L)}{1000mL} =\frac{382.14L}{1000}=0.38214L

0.38214 Liters.

4 0
3 years ago
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