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3 years ago

Is a Face Mask (the one you use if you go out during this pandemic) a homogeneous mixture?

Chemistry

2 answers:

mrs_skeptik [129]3 years ago

7 0

The answer is Yes it is

KATRIN_1 [288]3 years ago

6 0

Answer:

yes it is homogeneous mixture

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What is the percent composition of each element in the compound below: CrPO4

Paha777 [63]

Chromium phosphate pentahydrate

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3 years ago

What is the oxidation number of S In Li2SO4? A. +1 B. +2 C. +5 D. +6

-Dominant- [34]

Answer:

Explanation:

Li has an oxydation number of 1 2*1 = 2

O has an oxidation number of -2 4(-2) = -8

S has an oxydation number of x 1 * x = x

The oxidation number on the molecule is 0.

So here is the equation

2 - 8 + x = 0 Combine like terms

-6 + x = 0 Add 6 to both sides

-6 + 6 + x = 6 Combine like terms

x = 6

Sulphur in this case has an oxidation number of 6

3 0

3 years ago

Does the reaction between sodium

Masteriza [31]

The reaction between sodium

and chlorine is NaCl which is physical change

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4 years ago

Which order is correct in listing the bond lengths from shortest to longest? A. single, double, triple B. triple, double, single

Alex787 [66]

A I believe would be the answer

3 0

4 years ago

Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibri

melisa1 [442]

Answer:

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

Explanation:

The equilibrium constant will remain the same irrespective of the amount of reactant taken.

Let us calculate the equilibrium constant of the reaction.

Kc= $\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}$

Let us calculate the moles of each present at equilibrium

$moles=\frac{mass}{molarmass}$

molar mass of PCl₅=208

molar mass of PCl₃=137

molar mass of Cl₂=71

moles of PCl₅ = $\frac{mass}{molarmass}=\frac{4.13}{208}=0.02$

moles of PCl₃= $\frac{mass}{molarmass}=\frac{8.87}{137}=0.06$

moles of Cl₂ = $\frac{mass}{molarmass}=\frac{2.90}{71}=0.04$

the volume is 5 L

So concentration will be moles per unit volume

Putting values

Kc = $\frac{\frac{0.06}{5}\frac{0.04}{5}}{\frac{0.02}{5}}=0.024$

Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow

$PCl_{5}--->PCl_{3}+Cl_{2}$

Initial 0.02 0.06 0.04

Change -x +x +x

Equilibrium 0.02-x 0.06+x 0.04+x

Conc. (0.02-x)/2 (0.06+x)/2 (0.04+x)/2

Putting values

0.024 = $\frac{(0.06+x)(0.04+x)}{(0.02-x)2}$

Solving

$(0.024(2)(0.02-x)=(0.06+x)(0.04+x)$

$0.00096-0.048x=0.0024+x^{2}+0.1x$

$0.148x+x^{2}+0.00144=0$

x = -0.01

so the new moles of

PCl₅ = 0.02 + 0.01 =0.03

PCl₃ = 0.06-0.01 = 0.05

Cl₂ = 0.04-0.01 = 0.03

mass of each will be:

mass= moles X molar mass

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

5 0

4 years ago