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Step2247 [10]
3 years ago
15

How is mass calculated given density and volume Volume divided by volume, Sum of volume and mass, Density multiplied by volume,

Difference of volume and mass
Chemistry
2 answers:
Naddik [55]3 years ago
7 0
D = m/v
m = dv
v = m/d
So, the answer is density multiplied by volume.
Damm [24]3 years ago
3 0

C. Density multiplied by volume.

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HEY I NEED ALL ANSWERS RIGHT NOW IT'S MISSING AND MY TEACHER ARE GRADING THEM NOW YOU WILL BE MARKED BRAINIEST
emmainna [20.7K]

Molar mass of butane:-

  • 4(12)+10=58u

Moles of Butane:-

  • 100/58=1.7mol

#16

  • 2mols of Butane produce 8mol CO_2
  • 1mol of butane produces 4mol CO_2

Moles of CO_2

  • 4(1.7)=6.8mol

Mass of CO_2:-

  • 44(6.8)=299.2g

#17

  • 2mols of butane need 13mol O_2
  • 1mol of butane needs 6.5mol O_2

Moles of O_2

  • 6.5(1.7)=11.05mol

Mass of O_2

  • 11.05(32)=353.6g
7 0
2 years ago
2. What pressure is required to compress 196.0 L of air at 1.00 atmosphere into a cylinder
damaskus [11]

Answer:

<h2>7.54 atm </h2>

Explanation:

The required pressure can be found by using the formula for Boyle's law which is

P_2 =  \frac{P_1V_1}{V_2}  \\

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

From the question we have

P_2 =  \frac{1 \times 196}{26}  =  \frac{196}{26}   \\ = 7.538461...

We have the final answer as

<h3>7.54 atm </h3>

Hope this helps you

3 0
2 years ago
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
2 years ago
Describe the current model of the atom. Name the parts of this model and describe each part.
aivan3 [116]

Answer:

the electron cloud model is the current model of atom

7 0
3 years ago
Read 2 more answers
URGENT !! A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it h
Ede4ka [16]
<h3><u>Answer;</u></h3>

Empirical formula = C₂H₃O

Molecular formula = C₁₄H₂₁O₇

<h3><u>Explanation</u>;</h3>

Empirical formula

Moles of;

Carbon = 55.8 /12 = 4.65 moles

Hydrogen = 7.04/ 1 = 7.04 moles

Oxygen  = 37.16/ 16 = 2.3225 moles

We then get the mole ratio;

4.65/2.3225 = 2.0

7.04/2.3225 = 3.0

2.3225/2.3225 = 1.0

Therefore;

The empirical formula = <u>C₂H₃O</u>

Molecular formula;

(C2H3O)n = 301.35 g

(12 ×2 + 3× 1 + 16×1)n = 301.35

43n = 301.35

  n = 7

Therefore;

Molecular formula = (C2H3O)7

                             <u> = C₁₄H₂₁O₇</u>

6 0
3 years ago
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