All categories

3 years ago

Calculate the percent composition of Ca3P2

Chemistry

2 answers:

Afina-wow [57]3 years ago

8 0

Calculate the molar mass of Ca3P2 in grams per mole or search for a chemical formula or substance.

torisob [31]3 years ago

7 0

Answer:

Ca - 66%, P - 34%

Explanation:

So, this is the formula we can use to find the amount of each element:

Element count * Atomic mass = Mass

Plug in our elements for this:

Ca - 3*40.078=120.234

P - 2*30.973=61.946

Now, to find the percentage of mass, we must find total mass, and divide the two elements mass count by this total mass:

120.234+61.946=182.18

Now divide each element mass by the total mass:

Ca - 120.234/182.18=0.6599(Round to 0.65)

P - 61.946/182.18=0.34002(Round to 0.34)

Then multiply both numbers by 100 to get the percentage:

Ca - 65%

P - 34%

So these our your two answer!

Hope this helps!

You might be interested in

Need help on atoms and molecules

balu736 [363]

1-b

2-c

3-a

4-d

5-d

6-b

7-a

7 0

3 years ago

Calculate the mass of oxygen in 30 g of CH NH COOH?

Naya [18.7K]

Molar mass of CH2NH2COOH - 75

Given mass of CH2NH2COOH - 30

Moles of CH2NH2COOH = Given mass/ Molar mass

moles of CH2NH2COOH = 30/75 = 0.4 mol

One mole of CH2NH2COOH contains 32 gram of oxygen

0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen

Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!

7 0

2 years ago

This question in the screenshot

Morgarella [4.7K]

Mass of hydrate + crucible = 47.29 g

Mass of anhydrous salt = 2.7 g

Molar mass of anhydrous salt CuSO4 = 159.5 g

Given,

mass of empty crucible = 42.45 g

mass of hydrate salt= 4.84 g

mass of crucible after first heating = 46.1 g

mass of crucible after second heating= 45.153 g

mass of crucible after third heating= 45.15 g

so, as per the question we need to find...

Mass of hydrate + crucible = ? g

Mass of anhydrous salt = ? g

Molar mass of anhydrous salt CuSO4 = ? g

∴Mass of hydrate + crucible = 42.45 + 4.48 = 47.29 g

The given salt is in hydrate form, to remove water from this molecule we need to perform heating .

So we are taking the substance into the crucible as it is in less quantity.

Here, we performed heating 3 times and note the weight after every heating.

After this, assume that the water is totally evaporated and the remaining salt is in anhydrous form,

∴ Mass of anhydrous salt = 45.15 - 42.45 = 2.7 g

To find the molar mass of anhydrous salt of CuSO4,

atomic weight of Cu = 63.5 g

atomic weight of S = 32 g

atomic weight of O =16 g

∴ molar mass of anhydrous salt of CuSO4 = 63.5 + 32 + (16 ×3)

=159.5 gm

Learn more about molar mass here...

brainly.com/question/837939

#SPJ10

5 0

2 years ago

What human development period 2 years and 12 years called

nikdorinn [45]

Answer:

In these lessons, students become familiar with the four key periods of growth and human development: infancy (birth to 2 years old), early childhood (3 to 8 years old), middle childhood (9 to 11 years old), and adolescence (12 to 18 years old).

7 0

3 years ago

Read 2 more answers

Indicate the changes (increases, decreases, does not change) in its volume when the pressure undergoes the following changes at

Inga [223]

The question is missing information. Here is the complete question.

A gas at a pressure of 2.0 atm is in a closed container. Indicate the changes (if any) in its volume when the pressure undergoes the following changes at constant temperature and constant amount of gas. Match the words in the left with the column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.

1. The pressure increases to 6.0 atm. The volume ________

2. The pressure drops to 0.40 atm. The volume _________

3. The pressure remains at 2.0 atm. The volume _________

Answer: 1. Decreases

2. Increases

3. Does not change

Explanation: According to the Ideal Gas Law, Pressure, Volume and Temperature of an ideal gas is related, as the following: PV = nRT.

In this case, since temperature (T) and amount of gas (n) are constant, the Boyle's Law can be used.

The law states that the volume of a given gas, under the conditios of temperature and amount of it are constant, is inversely proportional to the applied pressure: P₁.V₁ = P₂.V₂

For case 1.)

Initial P (P₁) = 2

Initial V (V₁) = V

Final P (P₂) = 6

P₁.V₁ = P₂.V₂

2.V = 6.V₂

V₂ = 1/3V

When the pressure increases to 6 atm, volume decreases by 1/3.

For case 2.)

P₁ = 2

V₁ = V

P₂ = 0.4

2.V = 0.4V₂

V₂ = 5V

When pressure drops to 0.4 atm, volume increases by 5.

For case 3.)

Since there are no change in the pressure, the volume is the same from the beginning, so does not change.

8 0

3 years ago