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ehidna [41]
3 years ago
9

Automobile bodies contain significant amounts of iron. The iron is protected by the addition of zinc. This is called galvanizati

on, the protection of one metal the sacrifice of a more reactive metal. Write the balanced chemical equations using smallest whole number coefficients for the following processes. Do NOT include states-of-matter: a) The corrosion that is prevented by galvanization.
Chemistry
1 answer:
Oksana_A [137]3 years ago
6 0

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

Galvanization is defined as the process in which a protective layer of zinc is applied to iron or steel to prevent the metal from rusting.

Zinc prevents the oxidation of iron and acts as a reducing agent in the process.

The half reaction for the process follows:

<u>Oxidation half reaction:</u>  Zn\rightarrow Zn^{2+}+2e^-

<u>Reduction half reaction:</u>  Fe^{2+}+2e^-\rightarrow Fe

Net chemical equation:  Zn+Fe^{2+}\rightarrow Zn^{2+}+Fe

Hence, the balanced chemical equation is written above.

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Which of the following is the correct model of C6H14?
asambeis [7]

Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of options. However, the compound C₆H₁₄ is hexane. Hexane is a member of saturated hydrocarbons (homologous series) called alkanes (with the general formula CₙH₂ₙ₊₂). The structure for an hexane is shown below

     H   H   H   H   H   H

      I     I    I     I     I     I

H - C - C - C - C - C - C - H

      I     I     I    I      I    I

     H   H   H   H    H  H

which can also be written as

CH₃CH₂CH₂CH₂CH₂CH₃

3 0
2 years ago
Read 2 more answers
Find the number of moles in 6120 ions of NaCl . Round your answer to two decimal places. Input your answer as 1.03E23, which is
FrozenT [24]

Answer:

Explanation:

NaCl = Na⁺ + Cl⁻

6120 ions of NaCl will contain 3060 ions of Na⁺ and 3060 ions of Cl⁻ , forming 3060 molecules of NaCl .

6.02 x 10²³ molecules of NaCl = 1 mole

3060 molecules of NaCl = 3060 x 10⁻²³ / 6.02 moles

= 508.30 x 10⁻²³ moles

= 5.08 x 10⁻²¹ moles

Answer : 5.08E-21 moles  .

5 0
2 years ago
An aqueous solution of potassium sulfate is allowed to react with an aqueous solution of calcium nitrate. identify the solid in
Maksim231197 [3]
Ans: Calcium sulfate.

K2SO4 (aq) + Ca(NO3)2 (aq) ⇒ 2KNO3 (aq) + CaSO4 (s)

6 0
3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
2 years ago
Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a
svetoff [14.1K]

Answer:

The answer to your question is        P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

                      = 373°K

Formula

               (P + \frac{a}{v^{2}} )(v - b) = RT

Substitution

               (P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)

Simplify

               (P + 0.0094)(22.3829) = 30.586

Solve for P

                           P + 0.0094 = \frac{30.586}{22.3829}

                           P + 0.0094 = 1.366

                                 P = 1.336 - 0.0094

                                P = 1.357 atm

7 0
3 years ago
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