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Lunna [17]
3 years ago
5

When nitrogen is removed from a sample of air by the process of nitrogen fixation, what happens to the total pressure of the sam

ple?
a. It increases.
b. It stays the same.
c. It decreases.
Chemistry
1 answer:
Komok [63]3 years ago
8 0

The correct answer is "C. It Decreases".

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Which structure is part of the central nervous system?
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The spinal cord is part of the central nervous system

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3 years ago
Compare the volume of 14.1 g of helium to 14.1 g of argon gas (under identical conditions).
s2008m [1.1K]
The Volumes can be calculated from Masses by using following Formula,

                                        Density  =  Mass / Volume
Solving for Volume,
                                        Volume  =  Mass / Density

Mass of Both Gases  =  14.1 g

Density of Argon at S.T.P  =  1.784 g/L

Density of Helium at S.T.P  =  0.179 g/L

For Argon:
                                        Volume  =  14.1 g / 1.784 g/L

                                        Volume  =  7.90 L

For Helium:
                                        Volume  =  14.1 g / 0.179 g/L

                                        Volume  =  78.77 L
4 0
3 years ago
A sample of gas occupies 0.04 L at 150 K. What volume does that sample occupy at 300 K?
Natalka [10]
C.
hope this helps :)
8 0
3 years ago
Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um
Andre45 [30]

Answer:

\large \boxed{\text{0.0038 mol/L}}

Explanation:

1. Calculate the initial moles of acid and base

\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}

2. Calculate the moles remaining after the reaction

                   OH⁻     +     H₃O⁺ ⟶ 2H₂O

I/mol:      0.0053       0.005 00

C/mol:    -0.00500   -0.005 00

E/mol:      0.0003              0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}

8 0
3 years ago
What is simple displacement reaction???give examples.
VARVARA [1.3K]

Answer:

<h2>Hey rose why you deleted my comments</h2>

3 0
3 years ago
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