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Rufina [12.5K]
3 years ago
7

What is the percent by mass of oxygen in fe2o3

Chemistry
2 answers:
Effectus [21]3 years ago
5 0
To get the formula mass we have to add the atomic mass of each atom
(55.847 x 2) + ( 16 x 3) = 160
this means that 16 x 3 = 48 are the grams of O
% = 48 x 100 / 160 =30 %

hope this helped mark me brainliest
anygoal [31]3 years ago
4 0
30% is the answer. Good luck
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Sulfur hexafluoride, SF6, is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 2.02 moles o
miss Akunina [59]

Answer:

9.15 atm

Explanation:

Ideal gas equation of state PV=nRT

P in hPa, V in L, n in mol, R is a constant which is 83.1 hpa*L/mol*k, T in kelvin.

Plug in all the number, and we will get:

P*6.21=2.02*83.1*343

P =9271.6(in hpa)=9.15 atm

8 0
3 years ago
Can someone help me with this? (WILL GIVE BRAINLIEST)
iragen [17]

the answer is in the picture, btw the molar mass for the first one is wrong, it should be 77.98, and the final product is 2.32

6 0
3 years ago
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Mashutka [201]

Answer:

hi

Explanation:

7 0
4 years ago
According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. A 150-lb
julsineya [31]

Answer:

The density of the swimmer is 0.0342 lbm/in3.

This value makes sense as the density of the body is very similar to the water.

Explanation:

If the swimmers is floating, the weight of the swimmer must be equal to the upward buoyant force.

We can express the weight force as the product of density and volume of the swimmer.

Then

\rho_{swimmer}*V_{swimmer}=\rho_{water}*V_{water_displaced}\\\\\rho_{swimmer}*V_{swimmer}=\rho_{water}*0.95**V_{swimmer}\\\\\rho_{swimmer}=0.95*\rho_{water}\\\\\rho_{swimmer}=0.95*0.036 lbm/in3\\\\\rho_{swimmer}=0.0342lbm/in3

It makes sense as the density of the body is very similar to the water.

5 0
3 years ago
An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

3 0
3 years ago
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