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IceJOKER [234]
2 years ago
5

his is the chemical formula for chromium(III) nitrate: . Calculate the mass percent of oxygen in chromium(III) nitrate. Round yo

ur answer to the nearest percentage.
Chemistry
1 answer:
Alinara [238K]2 years ago
5 0

Answer:

\%\ Composition\ of\ iron=69.92\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

The formula for chromium(III) nitrate is Cr(NO_3)_3

Molar mass of chromium(III) nitrate = 238.011 g/mol

1 mole of chromium(III) nitrate contains 9 moles of oxygen

Molar mass of oxygen = 16 g/mol

So, Mass= Molar mass*Moles = 16*9 g = 144 g

\%\ Composition\ of\ iron=\frac{Mass_{iron}}{Total\ mass}\times 100

\%\ Composition\ of\ iron=\frac{144}{238.011}\times 100

\%\ Composition\ of\ iron=69.92\ \%

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you have two samples of gray powder both which are flammable these are powders the same substance explain why ​
nasty-shy [4]

The amount of matter stays the same between the substances,

7 0
2 years ago
Read 2 more answers
Based on your observations of 1-tetradecanol when it reformed a solid after melting, does 1-tetradecanol form a crystalline or a
Nitella [24]

Answer:

It is a crystalline solid.

It is a white crystalline solid that is practically insoluble in water, soluble in diethyl ether and slightly soluble in ethanol

Explanation:

The difference between crystalline and amorphous is how this chemical compound transmits light.

When a chemical material or compound is said to be crystalline, it is the opposite of what we imagine, since its color is opaque and does not allow light to pass through it, that is why this compound, being crystalline, is opaque white. and if you want to see through it you will not see the other way since it is not "transparent".

On the other hand, amorphous chemical materials or compounds are seen through them from one side to the other, they are considered "transparent" and do not refract any color from the color range of light. That is why they are not opaque either, nor do they have a particular color like white. A clear example of an amorphous structure is glass or crystal.

6 0
1 year ago
Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole
icang [17]

Answer:

Explanation has been given below.

Explanation:

  • Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
  • Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
  • First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
  • Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
  • Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
  • Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
3 0
2 years ago
When a 3.22 g sample of an unknown hydrate of sodium sulfate, na2so4 ⋅ h2o(s), is heated, h2o (molar mass 18 g) is driven off. T
Mariana [72]

The value of X is 10 hence the formula of unknown hydrate   sodium sulfate is  NaSO4.10 H20

calculation

step 1:find the moles of NaSO4 and the moles of H2O

moles= mass/molar mass

moles of Na2SO4=1.42÷142=0.01 moles

moles of H20=  mass of H2O/molar mass of H2O

mass of H2O= 3.22-1.42=1.8g

mole of H2O is therefore 1.8÷18=0.1 moles

step 2: find the mole ratio by dividing each mole by smallest number of mole (0.01)

that is Na2So4= 0.01/0.01 =1

            H2O= 0.1/0.01=10

6 0
1 year ago
An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial
andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state )  P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁  →  P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ =  n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2  = 2.25 atm

P₂= 2.25 atm

8 0
1 year ago
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