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vagabundo [1.1K]
2 years ago
13

Calculate the density of sulfur dioxide gas (SO2) at 147°C and 29.9 kPa.

Chemistry
1 answer:
Yuki888 [10]2 years ago
4 0

The density of the gas is obtained as  0.55 g/L.

We can find the density of the gas using the relation;

PM = dRT

P = pressure of the gas

M = molar mass of the gas

d = density of the gas

R = gas constant

T = temperature

Now;

P = 29.9 kPa

R = 8.314 L·kPa mol·K.

T =  147°C + 273 = 420 K

d = ?

M = 64 g/mol

Substituting values;

d = PM/RT

d =  29.9 kPa ×  64 g/mol/ 8.314 L·kPa mol-1K-1 420 K

d = 1913.6/3491.88

d = 0.55 g/L

Learn more about density: brainly.com/question/952755

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What is the formula of the compound formed between the potassium ion and the sulfide ion?
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The formula for K+ and S-2 is K2S because you need 2 potassium ions to balance out the sulfide ion
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Describe the composition of humus and why it is an effective organic fertilizer.
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Answer:

Humus, which ranges in colour from brown to black, consists of about 60 percent carbon, 6 percent nitrogen, and smaller amounts of phosphorus and sulfur. As humus decomposes, its components are changed into forms usable by plants.

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2 years ago
How many atoms are in a sample of 175 grams of sodium (Na)? The answer needs to be <br> a x 10^b
ValentinkaMS [17]
Atomic mass Sodium ( Na ) = 22.98 u.m.a

22.98 g ----------------- 6.02x10²³ atoms
175 g ------------------- ?? atoms

175 x ( 6.02x10²³) / 22.98 =

4.58x10²⁴ atoms of Na

hope this helps!

4 0
3 years ago
A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment,
nikitadnepr [17]

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

4 0
2 years ago
antimony has two naturally occuring isotopes, sb121sb121 and sb123sb123 . sb121sb121 has an atomic mass of 120.9038 u120.9038 u
Luda [366]

Considering the definition of atomic mass, isotopes and atomic mass of an element, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

<h3>Definition of atomic mass</h3>

The atomic mass is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

<h3>Definition of isotope</h3>

Isotopes are the chemical elements in which atomic numbers are the same, but the number of neutrons is different.

<h3>Definition of atomic mass</h3>

The atomic mass of an element is the weighted average mass of its natural isotopes.

This is, the atomic masses of elements are usually calculated as the weighted average of the masses of the different isotopes of each element, considering the relative abundance of each of them.

<h3>Percent natural abundance of each isotope</h3>

In this case, antimony has two naturally occuring isotopes, sb121 and sb123. You know:

  • sb121 has an atomic mass of 120.9038 u.
  • sb121 has a percent natural abundance of x.
  • sb123 has an atomic mass of 122.9042 u.
  • sb123 has a percent natural abundance of 1 -x (or, what is the same, the abundance is 100% - x%, since both isotopes form 100% of the element.)
  • Antimony has an average atomic mass of 121.7601 u

The average mass of antimony is expressed as:

121.7601 u= 120.9038 u x + 122.9042 u× (1 -x)

Solving:

121.7601 u= 120.9038 u x + 122.9042 u - 122.9042 u x

121.7601 u - 122.9042 u= 120.9038 u x - 122.9042 u x

(-1.1441 u)= (-2.0014) x

(-1.1441 u)÷ (-2.0014)= x

<u><em>0.5726= x or 57.26%</em></u>

So, 1 -x= 1- 0.5716 → <u><em>1-x= 0.4284 or 42.84%</em></u>

<u><em /></u>

Finally, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

Learn more about average atomic mass:

brainly.com/question/4923781

brainly.com/question/1826476

brainly.com/question/15230683

brainly.com/question/7955048

#SPJ1

5 0
1 year ago
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