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3 years ago

Which of the following indicates a strong base is present?

Chemistry

2 answers:

Olenka [21]3 years ago

8 0

Answer:

Explanation:

Founders education answer

oksian1 [2.3K]3 years ago

4 0

<h3>B</h3><h3>brainliest! These bases completely dissociate in solutions of 0.01 M or less. The other bases make solutions of 1.0 M and are 100% dissociated at that concentration. There are other strong bases than those listed, but they are not often.</h3>

Strong bases are bases which completely dissociate in water into the cation and OH- (hydroxide ion). The hydroxides of the Group I (alkali metals) and Group II (alkaline earth) metals usually are considered to be strong bases. These are classic Arrhenius bases. Here is a list of the most common strong bases.

LiOH - lithium hydroxide

NaOH - sodium hydroxide

KOH - potassium hydroxide

RbOH - rubidium hydroxide

CsOH - cesium hydroxide

*Ca(OH)2 - calcium hydroxide

*Sr(OH)2 - strontium hydroxide

*Ba(OH)2 - barium hydroxide

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A solution contains 45 grams of nitrogen gas and 40 grams of argon. What is the mole fraction of argon in this solution?

Katarina [22]

<span>0.38

You first calculate the total moles by dividing the grams by molecular weight:
45 g N2 / 28.02 g/mol = 1.6 mol N2
40 g Ar / 39.95 g/mol = 1.0 mol

Then you divide the moles of Ar by the total number of moles:
1.0 / (1.6 + 1.0) = 0.38 mol fraction</span>

8 0

3 years ago

How many moles of hydrogen gas can be produced if 0.78 moles of hydrochloric acid reacts with excess solid zinc according to the

White raven [17]

Answer:

The answer to your question is It will be formed 0.39 moles of H₂

Explanation:

Data

moles of H₂ = ?

moles of HCl = 0.78

moles of Zinc = excess

Balanced chemical reaction

2 HCl + Zn ⇒ 1 H₂ + ZnCl₂

Process

1.- Use proportions to solve this problem. Consider the coefficients of the balanced reaction.

2 moles of HCl ---------------------- 1 mol of H₂

0.78 moles of HCl ----------------- x

x = (0.78 x 1) / 2

- Simplification

x = 0.78 / 2

- Result

x = 0.39 moles of H₂

3 0

3 years ago

What is the effect of adding more water to the following reaction at equilibrium? CO2 + H2O H.CO.

LenaWriter [7]

Answer:If the pressure of a gaseous reaction mixture is changed the equilibrium will shift to minimise that change. If the pressure is increased the equilibrium will shift to favour a decrease in pressure.

Explanation:

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3 years ago

Rita wants to make some toast for breakfast. She puts a slice of bread in the toaster. After 3 min, Rita notices that the sides

fredd [130]

The answer is D

Because the bread cannot go back to its previous state, and because a new substance was formed. :)

6 0

3 years ago

Read 2 more answers

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(a) <em>Balanced equation including N_2 from air</em>

The balanced equation <em>ignoring</em> N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

<em>Including</em> N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) <em>Air:fuel mass ratio for 100 % combustion</em>

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago