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Salsk061 [2.6K]
4 years ago
13

M = mol/g solve for mol

Chemistry
1 answer:
Ganezh [65]4 years ago
6 0

Explanation:

M=mol/g

M×g=mol

mol=M×g

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what would be the final volume when 2.20 M solution is made from 25.0 mL of a 12.0 M solution? plzz show work
cupoosta [38]

Answer:

136.36 mL

Explanation:

Here we have to use the dilution formula

From C1V1= C2V2

Where;

C1= initial concentration of the solution= 12.0 M

C2= final concentration of the solution= 2.20 M

V1 = initial volume of the solution= 25.0 ml

V2= final volume of the solution= ?????

Then recall;

C1V1=C2V2

V2 = C1V1/C2

Substituting values from the parameters given;

V2= 12.0 × 25.0 / 2.20

V2= 136.36 mL

7 0
3 years ago
Which element is chemically similar to chlorine?
ad-work [718]

bromine

Explanation:

halogens are a group of elemnts simlar to eachother

flourine, chlorine, and bromine

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3 years ago
What chemical bond will S and H form
Over [174]

Answer:

hydrogen sulfide (H2S)

Explanation:

8 0
4 years ago
Question 3 of 5
rusak2 [61]

Answer:

A. More mass

C. Shorter distance between them

Explanation:

The two characteristics of a body experiencing greater gravitational force are that they have mass and a shorter distance between them.

This is conformity with Newton's law of universal gravitation.

The law states that "every object attracts one another with a force that is directly proportional to their masses and inversely proportional to the square of the distance between them".

This law implies that the more the mass of two bodies, the more the gravitational force of attraction. And that the shorter the square of the distance between them, the more the attraction.

7 0
3 years ago
If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th
qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

K_c=4.74\times 10^4

Concentration at equilibrium:

[COBr_2]=1.58\times 10^{-6}M

[Co]=2.78\times 10^{-3}M

[Br_2]=2.51\times 10^{-5}M

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0
3 years ago
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