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3 years ago

11

Air with less than 19.5% oxygen cannot support human life. How many moles of oxygen would the caisson if the percent of oxygen i

n the air dropped to 19.4%.

1 answer:

irinina [24]3 years ago

8 0

Are we adding the percent?

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Determine how much energy is needed to change 50 g of liquid water at 100°C to steam.

marin [14]

Answer:

$Q=113,000J$

Explanation:

Hello there!

In this case, since the vaporization process is carried out in order to turn a liquid into a gas due to the addition of heat, we can use the following heat equation involving the heat of vaporization of water or any other substance:

$Q=m*\Delta _{vap}H$

Thus, since this heat of vaporization for water is 2259.36 J/g, we plug in this amount to obtain the total energy for this process.

$Q=50*2259.36 J/g\\\\Q=113,000J$

Which is positive due to the necessity of heat.

Regards!

5 0

3 years ago

When matter is a gass, a lot of the time it is ____?

Zielflug [23.3K]

I’d say most likely air, oxygen, or carbon dioxide

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3 years ago

Read 2 more answers

Example 13:0.29 grams of a hydrocarbon with vapour density 29 when burnt completely in oxygen produce 448 ml of carbon dioxide a

Juliette [100K]

Answer:

Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10

6 0

2 years ago

Which statement best describes a chemical property of a mineral

alina1380 [7]

Answer:

The mineral forms gas bubbles when a sample is stored in water

6 0

2 years ago

ook at sample problem 18.12 in the 8th ed Silberberg book. Write a balanced chemical equation (salt hydrolysis). So acetate ion

vfiekz [6]

Answer:

Here's what I get

Explanation:

1. Write the chemical equation

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵

Let's rewrite the equation as

A⁻ + H₂O ⇌ HA + OH⁻

2. Calculate Kb

$K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}$

3. Set up an ICE table

A⁻ + H₂O ⇌ HA + OH⁻

I/mol·L⁻¹: 0.35 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.35-x x x

4. Solve for x

$\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}$

Check for negligibility,

$\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}$

5. Calculate the pOH

[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹

pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88

6. Calculate the pH.

pH + pOH = 14.00

pH + 4.88 = 14.00

pH = 9.12

Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.

3 0

3 years ago