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N76 [4]
3 years ago
8

What is the mass in grams of 2.21 mole of Ca3(PO4)4?

Chemistry
1 answer:
lesya [120]3 years ago
7 0

Answer:

1105.29 grams

Explanation:

If you calculate the molar mass of Ca3(PO4)4 you'll get 500.13. Molar mass is the mass of one mole of a substance. Multiply 500.1194 by 2.21 because we have 2.21 moles to get 1105.264 grams.

The molar mass of Ca is 40.078 grams, and there are 3 of them. So 40.078 x 3 is 120.234 grams.

The molar mass of P is 30.974 and there are 4 P's, so 30.974 x 4 is 123.896 grams.

The molar mass of O is 16 grams and there are 6 of them. Again, 16 x 16 is 256 grams.

Add up 120.234 + 123.896 + 256 to get the molar mass of Ca3(PO4)4, 500.13.

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The average atomic mass of Eu is 151.96 amu. There are only two naturally occurring isotopes of europium, Eu with a mass of 151.
earnstyle [38]

Answer:

The percentage abundance of Eu isotopes are 52 %  and 48 % .

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope,:

% = x %

Mass = 151.0 amu

For second isotope :

% = 100  - x  

Mass = 153.0 amu

Given, Average Mass = 151.96 amu

Thus,

151.96=\frac{x}{100}\times {151.0}+\frac{100-x}{100}\times {153.0}

Solving for x, we get that:

x = 52 %

<u>Thus percentage abundance of Eu isotopes are 52 %  and 48 % .</u>

7 0
3 years ago
Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 28.7 moles of FeO. Y
Alexxx [7]

Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

                            FeO        :           Al

                             2            :            3

                          28.7          :         3/2×28.7 = 43.05 mol

Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.

8 0
3 years ago
A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man
kap26 [50]
<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

  • Mass of NaBr sample is 11.97 g
  • % composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

  • Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

  • A sample of 11.97 g of NaBr contains 22.34% of Na by mass
  • Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

  • Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

                         = 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0
3 years ago
I NEED HELP ASAP!!!
Rus_ich [418]

Explanation:

To answer this question, we'll need to use the Ideal Gas Law:

p

V

=

n

R

T

,

where

p

is pressure,

V

is volume,

n

is the number of moles

R

is the gas constant, and

T

is temperature in Kelvin.

The question already gives us the values for

p

and

T

, because helium is at STP. This means that temperature is

273.15 K

and pressure is

1 atm

.

We also already know the gas constant. In our case, we'll use the value of

0.08206 L atm/K mol

since these units fit the units of our given values the best.

We can find the value for

n

by dividing the mass of helium gas by its molar mass:

n

=

number of moles

=

mass of sample

molar mass

=

6.00 g

4.00 g/mol

=

1.50 mol

Now, we can just plug all of these values in and solve for

V

:

p

V

=

n

R

T

V

=

n

R

T

p

=

1.50 mol

×

0.08206 L atm/K mol

×

273.15 K

1 atm

= 33.6 L

this is not the answer but it will help you

do by the formula it is on the answer

3 0
3 years ago
This is the process where fossil fuels, forests, or other carbon-containing substances are burned, adding more carbon dioxide to
sattari [20]
The process where fossil fuels, forests, or other carbon-containing substances are burned, addin more carbon dioxide to the air is the combustion.

Some examples of combustion are:

Fossil fuel:

Carbon + O2

C + O2 -> CO2

Forests (wood)

Wood = cellulose = [C6H10O5]n

[C6H10O5]n + 6nO2 = 6n CO2 + 5n H2O

So, in general the combustion of organic matter produces CO2 and water.


4 0
3 years ago
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