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Marizza181 [45]
3 years ago
15

Hahajhjekajsjiekenkwlaoa

Chemistry
2 answers:
kodGreya [7K]3 years ago
7 0

Answer:

jeg;djb;dfjbejfnvjkndjv

Explanation:

i can not read it clearly

Anestetic [448]3 years ago
5 0

Answer:

.

Explanation:

I dont know if its correct correct but i think most are.

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What is the concentration of a 300 mL solution containing 0.27 mol of glucose?
Kryger [21]

Answer: B

Explanation:  concentratio = amoiunt of moles / volume

n = c/V = 0.27 mol / 0.30 l = 0.900 M

3 0
3 years ago
What is the chemical formula of iron (II) sulfide would be: (EO S = +2 and EO Fe = +2). please
Anni [7]

Answer:

FeS

Explanation:

The reason why their is no subscripts is because as put in the question, the charges are equivalent. As you can see, +2 and -2 equal each other out.

Note: S has a charge of -2, not +2. (Look at oxidation rules.)

7 0
3 years ago
Which property is the same for a 50 mL sample of liquid water and a 100 mL sample of liquid water?
Brums [2.3K]

Answer: The density because it stays the same with any same liquids or solids or gas with whatever amount of it is added. It only changes if the matter of state is different or there is any environmental change.

4 0
2 years ago
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and
N76 [4]

Answer:

Oxidizing agent - CrO4^2-

Reducing agent- N2O

Explanation:

Let us look at the equation closely;

CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]

The reduction half equation is;

CrO4^2- (aq) + 3e -------->Cr^3+ (aq)

Oxidation half equation is;

3N2O(g) ------>3 NO(g) +3 e

Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.

8 0
3 years ago
The element X has three naturally occurring isotopes. The masses (amu) and % abundances of the isotopes are given in the table b
denis-greek [22]

Answer:

220.42098 amu

Explanation:

(220 .9 X  .7422) + (220 X .0.1278) + (218.1 X 0.13) =     220.42098 amu

These are weighted averages.

So, we will take mass of one and multiply by abundance percentage that is provided and add them together.

In order to calculate the  average atomic mass, we have to convert the percentages of abundance to decimals. So, you get

(220 .9 X  .7422) + (220 X .0.1278) + (218.1 X 0.13) =     220.42098 amu

6 0
3 years ago
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