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aliya0001 [1]
3 years ago
5

How many moles are found in a 350.mL solution of 13.5M HNO3

Chemistry
1 answer:
tiny-mole [99]3 years ago
5 0

Answer: Moles are found in a 350.mL solution of 13.5M HNO_{3} are 4.725 mol.

Explanation:

Given: Volume = 350 mL (1 mL = 0.001 L) = 0.350 L

Molarity = 13.5 M

Molarity is the number of moles of a substance present in a liter of solution.

Therefore, moles of HNO_{3} present in the given solution are calculated as follows.

Molarity = \frac{moles}{Volume (in L)}\\13.5 M = \frac{moles}{0.350 L}\\moles = 4.725 mol

Thus, we can conclude that moles found in a 350.mL solution of 13.5M HNO_{3} are 4.725 mol.

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insert the missing coefficients in the following partially balanced chemical equations S+6HNO=HSO4+NO+H2O​
lianna [129]

Answer:

S + 6HNO₃ ⟶ H₂SO₄ +  6NO₂ + 2H₂O

Step-by-step explanation:

S + 6HNO₃ ⟶ H₂SO₄ +  NO₂ + H₂O  

One way is to balance this equation is the <em>oxidation number method</em>.

Step 1. Calculate the oxidation number of every atom:

S⁰ + H⁺¹N⁺⁵O₃⁻² ⟶ H₂⁺¹S⁺⁶O₄⁻² + N⁺⁴O₂⁻² + H₂⁺¹O⁻²

Step 2. Identify the changes in oxidation number:

S: +0 ⟶ +6; Change = +6

N: +5 ⟶ +4; Change =   -1

Step 3. Equalize the changes in oxidation number.

You need 6 atoms of N for every 1 atom of S. This gives us total changes of -6 and + 6.

Step 4. Insert coefficients to get these numbers.

1S + 6HNO₃ ⟶ 1H₂SO₄ +  6NO₂ + H₂O

Step 5. Balance O.

We have fixed 18 O on the left and 16 O on the right. We need 2 more O on the right. Put a 2 in front of H₂O.

1S + 6HNO₃ ⟶ 1H₂SO₄ +  6NO₂ + 2H₂O

Every formula now has a coefficient. The equation should be balanced.

Step 6. Check that all atoms balance.

<u>Atom</u>  <u>Left-hand side</u>  <u>Right-hand side</u>  

 S                  1                           1

 H                 6                          6

 N                 6                          6

 O               18                          18

The balanced equation is

S + 6HNO₃ ⟶ H₂SO₄ +  6NO₂ + 2H₂O

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Explanation:

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TRUE or FALSE: Broken wires or water can cause a long circuit. O True O False​
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Action Balanced Equation Particle View (center of
katrin2010 [14]

Answer:

N₂ + 3H₂  →  2NH₃

2H₂O →  2H₂ + O₂

CH₄ + 2O₂  →  CO₂ + 2H₂O

Explanation:

A) Make ammonia:

Chemical equation:

N₂ + H₂  →  NH₃

Balanced chemical equation:

N₂ + 3H₂  →  2NH₃

Step one:

N₂   +   H₂      →     NH₃

N = 2                    N = 1

H = 2                    H = 3

Step 2:

N₂   +   3H₂      →          2NH₃

N = 2                              N = 1

H = 2×3 = 6                    H = 3×2 = 6

Step 3:

N₂   +   3H₂      →          2NH₃

N = 2                              N = 1×2 = 2

H = 2×3 = 6                    H = 3×2 = 6

B) separate water:

Chemical equation:

H₂O →  H₂ + O₂

Balanced chemical equation:

2H₂O →  2H₂ + O₂

Step one:

H₂O        →        H₂ + O₂

H = 2                    H = 2

O = 1                     O = 2

Step 2:

2H₂O                →        H₂ + O₂

H = 2×2 = 4                  H = 2

O = 1×2 =2                     O = 2

Step 3:

2H₂O                →        2H₂ + O₂

H = 2×2 = 4                  H = 2×2 = 4  

O = 1×2 =2                     O = 2

C) Combust methane

Chemical equation:

CH₄ + O₂  →  CO₂ + H₂O

Balanced chemical equation:

CH₄ + 2O₂  →  CO₂ + 2H₂O

Step one:

CH₄ + O₂  →  CO₂ + H₂O

C = 1                    C = 1

H = 4                   H = 2

O = 2                    O = 3

Step 2:

CH₄ + 2O₂  →  CO₂ + H₂O

C = 1                     C = 1

H = 4                    H = 2

O = 2×2 = 4          O = 3

Step 3:

CH₄ + 2O₂  →  CO₂ + 2H₂O

C = 1                    C = 1

H = 4                   H = 2×2 = 4  

O = 2×2 = 4         O = 2 + 2 = 4

5 0
3 years ago
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