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kupik [55]
3 years ago
11

Rolls of foil are 304 mmmm wide and 0.014 mmmm thick. (The density of foil is 2.7 g/cm3g/cm3 .) What maximum length of foil can

be made from 1.06 kgkg of foil?
Chemistry
1 answer:
ratelena [41]3 years ago
3 0

Answer:

92.2 m

Explanation:

Given that:=

The breadth = 304 mm

Height = 0.014 mm

Let Length = x mm

Volume = Length\times breadth\times height

Thus,

Volume = 304\times 0.014\times x\ mm^3=4.256x\ mm^3

Also, 1 mm³ = 0.001 cm³

So, volume = 0.004256 cm³

Given that density = 2.7 g/cm³

Mass = 1.06 kg = 1060 g

So,

Volume=\frac{Mass}{Density}=\frac{1060}{2.7}\ cm^3=392.59\ cm^3

So,

0.004256*x = 392.59

x = 92243.89 mm

Length of foil = 92243.89 mm = 92.2 m

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Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . If of water i
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Answer:

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Explanation:

Balanced Chemical Equation:

HCl + NaOH = NaCl + H2O

We are Given:

Mass of H2O = 9.17 g

Mass of HCl = 21.1 g

Mass of NaOH = 43.6 g

First, calculate the moles of both HCl and NaOH:

Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles

Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles

Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:

Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles

Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles

From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:

Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g

% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%

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Please help me equalize: PbO2 + MnSO4 + HNO3 = HMnO4 + PbSO4 + Pb(NO3)2 + H2O
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Answer:

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O

Explanation:

PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O

It will be easiest to balance this equation by the ion-electron method.

1. Write the ionic equation

PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O

2. Eliminate H⁺, H₂O, and spectator ions

PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺  

3. Separate the skeleton equation into two half-reactions.

PbO₂  ⟶ Pb²⁺  

Mn²⁺ ⟶ MnO₄⁻

4. Balance all atoms other than H and O

Done

5. Balance O by adding water molecules to the deficient side

           PbO₂  ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻

6. Balance H by adding H⁺ ions to the deficient side.

  PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺

7. Balance charge by adding electrons to the deficient side.

PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O

     Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-

 

8. Multiply each half-reaction by a number to equalize the electrons transferred.

5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]

     2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]

9. Add the two half-reactions.

                          5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O

<u>                                    2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻                    </u>

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

10. Cancel species that occur on each side of the equation

5PbO₂ + 2Mn² + <u>8H₂O</u> + <u>20H⁺</u> + <u>10e⁻</u> ⟶ 5Pb²⁺ + 2MnO₄⁻ + <u>10H₂O</u> + <u>16H⁺</u> + <u>10e⁻ </u>

becomes

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

11. Add the missing spectator ions

5PbO₂ + 2Mn²⁺    + 4H⁺                              ⟶            5Pb²⁺   + 2MnO₄⁻ + 2H₂O

            + 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻       +2SO₄²⁻ + 6NO₃⁻ + 2H⁺

becomes

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

12. Check that all atoms are balanced.

\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Pb} & 5 & 5\\\text{O} & 36 & 36\\\text{S} & 2 & 2\\\text{H} & 6 & 6\\\text{N} & 6 & 6\\\end{array}

Everything checks. The balanced equation is

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

7 0
3 years ago
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