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ella [17]
3 years ago
12

The founder of Islam was _____. Abbas ibn Abd al-Muttalib Muhammad Abu Bakr Khosrau

Chemistry
2 answers:
Grace [21]3 years ago
5 0
The founder of Islam was Muhammad
-Corey
raketka [301]3 years ago
3 0
Muhammad was the founder of Islam
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Calculate the molarity of (3.25 mol of LiCl in 2.78 L solution
alisha [4.7K]

\text{Morality}= \dfrac{\text{Number of moles}}{\text{Volume}}=\dfrac{3.25}{2.78}=1.169 ~mol/L

8 0
2 years ago
Calculate the pH of a 0.10 M HCN solution that is 0.0070% ionized.
Anastaziya [24]

Answer:

D) 5.15

Explanation:

Step 1: Write the equation for the dissociation of HCN

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Step 2: Calculate [H⁺] at equilibrium

The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.

α% = [H⁺]eq / [HCN]₀ × 100%

[H⁺]eq = α%/100% × [HCN]₀

[H⁺]eq = 0.0070%/100% × 0.10 M

[H⁺]eq = 7.0 × 10⁻⁶ M

Step 3: Calculate the pH

pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15

7 0
3 years ago
What are blocks of time in the geological time scale called?
Katena32 [7]
B is the answer , i got you
7 0
3 years ago
Read 2 more answers
The following reaction is at equilibrium in a sealed container.
ale4655 [162]

Answer:

D.Lowering the temperature is the best option.

Explanation:

The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.

In the reaction below for example;

A + B <==>C+D

If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?

Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?

If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.

3 0
3 years ago
Draw the alkene formed when 1-heptyne is treated with hbr in the presence of peroxide.
Nimfa-mama [501]
<h2>Heptene formed is -</h2><h2>CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr</h2>

Explanation:

The two possibilities when the peroxide is not present

  • CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-C≡CH +HBr → CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_{2}

  • CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_2 + HBr →CH_3-CH_2-CH_2-CH_2-CH_2-CBr_2-CH_3

In presence peroxide,

CH_3-CH_2-CH_2-CH_2-CH_2-C≡CH+ HBr →CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr

  • When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
  • This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
  • One mole of HBr adds to one mole of 1-heptane.
  • The structure of heptene formed is -

CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr

5 0
3 years ago
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