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natka813 [3]
3 years ago
8

What is the volume of 2.5 moles of nitrogen gas (N2)at standard temperature and pressure (STP)?​

Chemistry
1 answer:
gregori [183]3 years ago
8 0

Answer:

1 mole of gas = 22.4L

2.5 moles of gas takes up = ( 22.4 L/ 1 mole ) x 2.5 mole

= 56 L

Explanation:

Please mark brainliest and have a great day!

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Baking soda (NaHCO3) and vinegar (HC2H3O2) react to form sodium acetate, water, and carbon dioxide. If 42.00 g of baking soda re
Setler [38]

Answer:

0.5 mole of CO₂.

Explanation:

We'll begin by calculating the number of mole in 42 g of baking soda (NaHCO₃). This can be obtained as follow:

Mass of NaHCO₃ = 42 g

Molar mass of NaHCO₃ = 23 + 1 + 12 + (16×3)

= 23 + 1 + 12 + 48

= 84 g/mol

Mole of NaHCO₃ =?

Mole = mass / molar mass

Mole of NaHCO₃ = 42/84

Mole of NaHCO₃ = 0.5 mole

Next, balanced equation for the reaction. This is given below:

NaHCO₃ + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O + CO₂

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂

Finally, we shall determine the number of mole of CO₂ produced by the reaction of 42 g (i.e 0.5 mole) of NaHCO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂.

Therefore, 0.5 mole of NaHCO₃ will also react to produce 0.5 mole of CO₂.

Thus, 0.5 mole of CO₂ was obtained from the reaction.

7 0
2 years ago
Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
4 0
3 years ago
How many moles of O2 are contained in 32 grams of O2?
Radda [10]

Answer:

2 moles

Explanation:

In one mole of O2 there are 16 grams. So in 2 moles there are 32 grams

7 0
2 years ago
How are oxidation-reduction reactions related to how you use energy?
Tems11 [23]

Answer:

In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. They also flow by a wire

Explanation:

8 0
2 years ago
Learning Task 1: Choose the correct answer. Write your answers on your
dybincka [34]

panjang banget gua kaga ngerti

7 0
2 years ago
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