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MissTica
3 years ago
7

Humans affect the levels of air pollution by using natural resources and

Chemistry
1 answer:
Radda [10]3 years ago
5 0
Chemicals that’s and ghebreziqbiher
You might be interested in
What does the roman numeral stand for in copper(1) oxide should it not be copper(II) oxide
photoshop1234 [79]

Answer:

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

Explanation:

Roman numeral is used to indicate the oxidation number of an element in a compound.

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

This can be seen from the following illustration:

copper(I) oxide => Cu₂O

Oxidation number of O = –2

Oxidation number of Cu₂O = 0

Oxidation number of Cu =?

Cu₂O = 0

2Cu + O = 0

2Cu – 2 = 0

Collect like terms

2Cu = 0 + 2

2Cu = 2

Divide both side by 2

Cu = 2/2

Cu = 1

Thus, we can see that the oxidation number of Cu in Cu₂O is 1. Hence the name of Cu₂O is copper(I) oxide indicating that the oxidation number of of copper (Cu) in the compound is 1.

For copper(II) oxide, we shall determine the oxidation number of Cu. This can be obtained as follow:

copper(II) oxide, CuO => CuO

Oxidation number of O = –2

Oxidation number of CuO = 0

Oxidation number of Cu =?

CuO = 0

Cu + O = 0

Cu – 2 = 0

Collect like terms

Cu = 0 + 2

Cu = 2

Thus, the oxidation number of Cu in CuO is 2. Hence the name of CuO is copper(II) oxide indicating that the oxidation number of of copper (Cu) in the compound is 2.

From the above illustrations,

We can see that the roman numeral in both copper(I) oxide, Cu₂O and copper(II) oxide, CuO are different because the oxidation number of Cu in both cases are different.

3 0
3 years ago
The molar mass of NH3 is 17.03 g/mol. The molar mass of H2 is 2.0158 g/mol. In a particular reaction, 0.575 g of NH3 forms. What
Arlecino [84]

The balanced chemical reaction is:

N2 + 3H2 = 2NH3

 

We are given the amount of ammonia formed from the reaction. This is where we start our calculations.

 

0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2

3 0
3 years ago
Read 2 more answers
4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.
slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ\frac{1mol}{102g} = <em>0,0980 moles</em>

And 10,0g of HCl are:

10,0 gₓ\frac{1mol}{36,5g} = <em>0,274 moles</em>

<em />

For a total reaction of 0,274 moles of HCl you need:

0,274×\frac{1molesAl_{2}O_3}{6 mole HCl} = <em>0,0457 moles of Al₂O₃</em>

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl ×\frac{2 moles AlCl_{3}}{6 moles HCl} × 133\frac{g}{mol} = <em>12,1 g of AlCl₃</em>

<em />

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles

And its mass is:

0,0523 molesₓ\frac{102g}{1mol} = <em>5,33 g of Al₂O₃ </em>

<em />

I hope it helps!

7 0
3 years ago
For an object to remain at rest, which of the following must be true?
zubka84 [21]

Answer:

the forces on it are balanced

Explanation:

7 0
3 years ago
Read 2 more answers
If an aqueous solution of ___________ is added to a mixture of barium nitrate and silver nitrate, the barium ion will precipitat
kkurt [141]

Answer:

sodium fluoride

Explanation:

When we add sodium fluoride to a solution of a given mixture, we get barium fluoride from Ba, which would be an insoluble salt, and silver fluoride from Ag, which would be a soluble salt. 

The solubility rule will be used to determine the barium salt that forms as a precipitate and leaves Ag+ salt in the solution.

From the following equations, we will see that the precipitate is formed in Ba but Ag remains dissolved in the solution.

Ba(NO_3)_2{(aq)} + 2NaF_{(aq)} \to BaF_{2(s)} +2NaNo_{3(aq)}

AgNo_{3(aq)} +NaF_{(aq)} \to AgF_{(aq)} + NaNO_{(aq)}

4 0
3 years ago
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