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Yuri [45]
3 years ago
14

Which of the following statements concerning the density of a gas is true?

Chemistry
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:

can u give us the options

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How many moles are in a 62.5-g sample of potassium nitrate (KNO3)?
Reil [10]

Answer:25.3 g of KNO₃ contain 0.25 moles.

Explanation:

7 0
2 years ago
Because of the mud in the muddy water eventually settles out the muddy water is a ____________.
sp2606 [1]

it should be B. solution; a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).

witch basically means it sorts it self out when left a lone, like soap and water when not touched or the oil in the dressings you get at the store. hope this helped :D


6 0
3 years ago
Read 2 more answers
True or false the fission of a nucleus is accompanied by the release of very little energy
4vir4ik [10]

Answer:

false

Explanation:

fission is lots of energy

7 0
3 years ago
Caffeine, a stimulant found in coffee and soda, hasthe mass percent composition: C. 49.48%, H, 5.19%. N. 28.85% 0. 16.48% The mo
borishaifa [10]

We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

C_4H_5O_1N_2

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

C_8H_{10}O_2N_4

8 0
1 year ago
skid marks about 6.7 km long. If the car had an initial speed of 425 km/h, estimate the coefficient of kinetic friction. The acc
Zanzabum

Answer:

Explanation: F=umg where u is coefficient of kinetic friction.

But F=ma.

a = u^2/2S

U= 425km/hr = (425*1000)/3600 =118m/s

S= 6.7km = 6700m

a = 118*118/ (2*6700) = 1.04m/s

u = 1.04/9.8 =0.11

8 0
3 years ago
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