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2 years ago

Find the percent composition of each element in KMnO4

1 answer:

6 0

The way you want to find the percent composition would be by breaking down the problem like so:

K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999

Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO

Then you want to take each molar mass and then divide it 110.035 and multiply by 100

Ex. K = 39.098/ 110.035 and the multiply what you get by a 100

You do this for the other elements as well good luck!

You might be interested in

Perform the following operation

Yanka [14]

Taking into account the scientific notation, the result of the sum is 10.84300×10³.

Scientific notation

First, remember that scientific notation is a quick way to represent a number using powers of base ten.

The numbers are written as a product:

a×10ⁿ

where:

a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.

Sum in scientific notation

You want to add two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 3.

Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 9.7300×10²= 0.97300×10³

Taking the quantities to the same exponent, all you have to do is add what was previously called the number "a". In this case:

0.97300×10³ + 9.8700×10³= (0.97300+ 9.8700)×10³= 10.84300×10³

Finally, the result of the sum is 10.84300×10³.

Learn more:

3 0

1 year ago

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is $CO3^{-2}$ , and the ion nitrate is $NO3^{-}$ .

Sodium is in group 1, so it must lose one electron to be stable, and be the cation $Na^{+}$ . Silver has only one electron too, so the cation will be $Ag^{+}$ .

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0

2 years ago

2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s

Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.

In order to prepare 10 mL, 10 μM; 4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM; 6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM; 8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

= 2 mL

2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

6 0

2 years ago

Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas? a. Mg b. Li c. Al d. Pb

Stells [14]

I believe the answer is A) Mg.

5 0

2 years ago

Read 2 more answers

Which equation is used to help form the combined gas law? eP, V, P, V, т.

alexdok [17]

Answer:

The combined gas law is formulated from PV/T =K.

Explanation:

The combined gas law comprises of Boyle's law, Charles's law and Gay lusaac's law. This laws were not discovered but simply put together considering other cases of ideal gas law. It states that if the amount of gas is left unchanged, the ratio between the pressure, volume, and temperature is constant.

7 0

2 years ago