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3 years ago

What is the mass of 2.5 moles of hydrogen fluoride gas HF

Chemistry

1 answer:

Nady [450]3 years ago

7 0

Answer:

You will get 5.0 g of hydrogen.

Explanation:

As with any stoichiometry problem, we start with the balanced equation.

2HF

→

SnF

Moles of H

2.5

mol Sn

1 mol H

mol Sn

2.5 mol H

Mass of H

2.5

mol H

2.016 g H

mol H

5.0 g H

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4 years ago

What are the most enzymes in the body?

Alex787 [66]

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4 years ago

How many liters are there in 3.0 moles of CO2 at STP?

Lerok [7]

There are 67.2 liters of CO2 at STP

<h3>Further explanation</h3>

Given

3 mole of CO2

Required

Volume of CO2 at STP

Solution

Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

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8 0

3 years ago

Basic amino acids are ________ (positive, negative) at pH 7 and acidic R group amino acids are ________ (positive, negative) at

Pani-rosa [81]

Answer:

Positive, Negative

Explanation:

The charge of an amino acid is dependent on pH values of their pKa's. At pH 7, Basic amino acid will be positive because they have pH values below their pKa's while an R group amino acid will be negative

7 0

3 years ago

References Use the References to access important values if needed for this question A student ran the following reaction in the

ANTONII [103]

Answer:

19.27

Explanation:

Some values are corrected from correct source.Thus,

Moles of SO₂ = 8.19x10⁻² moles

Moles of O₂ = 8.10x10⁻² moles

Volume = 1 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Concentration of SO₂ = 8.19x10² M

Concentration of O₂ = 8.10x10 M

Considering the ICE table for the equilibrium as:

2SO₂ (g) + O₂ (g) ⇔ 2SO₃ (g)

t = o 8.19x10⁻² 8.19x10⁻²

t = eq -2x -x 2x

--------------------------------------------- --------------------------

neq: 8.19x10⁻² -2x 8.19x10⁻² -x 2x

Given:

Equilibrium concentration of O₂ = 5.98x10⁻² M = 8.19x10⁻² -x

Thus, x = 0.0212 M

[SO₂] = 8.19x10⁻² - 2*0.0212 = 0.0395 M

[SO₃] = 2*0.0212 = 0.0424 M

The expression for the equilibrium constant is:

$K_c=\frac {[SO_3]^2}{[SO_2]^2[O_2]}$

$K_c=\frac{{0.0424}^2}{{0.0395}^2\times 0.0598}$

K = 19.27

4 0

3 years ago