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Thepotemich [5.8K]
10 months ago
11

Swift has a balloon of gas at 64psi. The volume of the balloon is 6.958L. If the volume of the balloon is reduced to 3L. what wi

ll the new pressure be in psi? (Assume the temperature is constant)
Chemistry
1 answer:
aniked [119]10 months ago
3 0

We could use the Boyle's law for gases.

This law states that:

P_1V_1=P_2V_2

Which means that at a constant temperature, as pressure increases, volume decreases. (P1 and V1 are the initial pressure and volume and P2 and V2 are the final pressure and volume respectively)

We could replace the values of the problem to get:

\begin{gathered} P_1=64psi \\ V_1=6.958L \\ P_2=? \\ V_2=3L \end{gathered}

Now, replacing in the equation:

\begin{gathered} 64psi\cdot6.958L=P_2\cdot3L \\ P_2=\frac{64psi\cdot6.958L}{3L}=148.44psi \end{gathered}

Therefore, the new pressure will be 148.44psi.

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Answer:

1. MG

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4. NA

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2 years ago
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How many atoms of cobalt are in 4 moles of cobalt?
Anna [14]
<h3>Answer:</h3>

2 × 10²⁴ atoms Co

<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
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

4 mol Co (Cobalt)

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 4 \ mol \ Co(\frac{6.022 \cdot 10^{23} \ atoms \ Co}{1 \ mol \ Co})
  2. Multiply/Divide:                 \displaystyle 2.4088 \cdot 10^{24} \ atoms \ Co

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

2.4088 × 10²⁴ atoms Co ≈ 2 × 10²⁴ atoms Co

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2 years ago
Help Asap plz!!!!!!!!
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I think C. Mutualism.

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3 years ago
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(b) What is the pOH of 1.35x10⁻³ M HCl? Is the solution neutral, acidic, or basic?
nika2105 [10]

The pOH scale is similar to the pH scale in that a pOH of 7 is indicative of a neutral solution. A basic solution has a pOH less than 7, while an acidic solution has a pOH of greater than 7

<h3>What is pOH scale?</h3>

Based on the relative concentration of hydrogen and hydroxyl ions, the pH and pOH scales are used primarily in chemistry, biology, and soil science to describe and measure the acidity and basicity of neutral, acidic, and basic solutions in water.

The terms pH and pOH scale are frequently used and crucial for determining the chemical, biological, and soil health by converting the values of hydrogen or hydroxyl ion concentrations or by utilizing pH paper and meters.

The pH scale or chart, which measures the concentration of hydrogen and hydroxyl ions in acidic and alkaline solutions, was initially developed by Sorensen.

To learn more about pOH scale from the given link:

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5 0
1 year ago
Hydrogen can be extracted from natural gas according to the following equilibrium.
antiseptic1488 [7]

Answer:

2.16x10⁻²

Explanation:

First, let's find out the molar concentrations of the reactants. The molar mass of CH4 is 16 g/mol, and of CO2 is 44 g/mol. The number of moles is the mass divided by the molar mass:

nCH4 = 24.0/16 = 1.5 moles

nCO2 = 88.0/44 = 2 moles

The concentration is the number of moles divded by the volume, thus:

[CH4] = 1.5/1 = 1.50 M

[CO2] = 2/1 = 2.00 M

For the equilibrium reaction, let's do an equilibrium chart:

CH4(g) + CO2(g) ⇄ 2CO(g) + 2H2(g)

1.50 2.00 0 0 Initial

-x -x +2x +2x Reacts (stoichiometry is 1:1:2:2)

1.50-x 2.00-x 2x 2x Equilibrium

As sateted in problem, [CH4] = 2.70*[CO]

1.50 - x = 2.70*2x

1.50 - x = 5.4x

6.4x = 1.50

x = 0.2344

Thus, at equilibrium:

[CH4] = 1.50 - 0.2344 = 1.2656 M

[CO2] = 2.00 - 0.2344 = 1.7656 M

[CO] = 2*0.2344 = 0.4688 M

[H2] = 2*0.2344 = 0.4688 M

The equilibrium constant is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients.

K = ([CO]²*[H2]²)/([CH4]*[CO2])

K = (0.4688²*0.4688²)/(1.2656*1.7656)

K = 0.0483/2.2345

K = 2.16x10⁻²

7 0
2 years ago
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