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Travka [436]
3 years ago
14

In a particular electroplating process, the metal being plated has a +4 charge. If 861.8 C of charge pass through the cell, how

many moles of metal should be plated? Useful information: F = 96,500 C/mol e- Provide your response to four digits after the decimal.
Chemistry
1 answer:
Xelga [282]3 years ago
3 0

Answer: 2.2326\times 10^{-3} moles

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

The metal being plated has a +4 charge, thus the equation will be:

M^{4+}+4e^-\rightarrow M

4\times 96500C of electricity deposits = 1 mole of metal

Thus 861.8 C of electricity deposits =\frac{1}{4\times 96500}\times 861.8=2.2326\times 10^{-3} moles of metal

Thus 2.2326\times 10^{-3} moles of metal should be plated

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A. is the correct point.

Explanation:

This is true because no matter how many mL of water is added, the solution only gets more height; the concentration in everything else stays the same, and water doesn't have any concentration. Very confusing, I know. Good luck!

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3 years ago
Which best describes an element?
Aleonysh [2.5K]

I think the answer is a pure substance

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3 years ago
How many milliliters of .085 m naoh are required to titrate 25 ml of .072 m hbr to the equivalence point?
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The ML  of 0.85  m NaOH    required   to  titrate  25 ml of  0.72m hbr  to  the  equivalence  point  is calculated  as  follows

calculate  the moles  of HBr used

moles  = molarity  x  volume

25  x0.072/1000=  0.0018 moles


write the  equation  for  reaction

NaOH + HBr = NaBr  +  H2O
from   reacting   equation the  mole ratio  between  NaOH  to  HBr  is  1:1  therefore  the  moles of  NaOH  =  0.0018 moles

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3 years ago
Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.
JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

  • H: 1.008;
  • Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

  • Hydrogen gas \rm H_2, and
  • Magnesium chloride, which is a salt.

The chemical equation will be something like

\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}],

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of \rm H_2 that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be \rm MgCl_2.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq).

Balance the equation. \rm MgCl_2 contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq).

The number of \rm Mg and \rm Cl atoms shall be the same on both sides. Therefore

\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq).

The number of \rm H atoms shall also conserve. Hence the equation:

\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq).

How many moles of HCl are available?

M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}.

\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol.

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of \rm HCl is 2 while the coefficient in front of \rm H_2 is 1. In other words, it will take two moles of \rm HCl to produce one mole of \rm H_2. \rm 4.52576\;mol of \rm HCl will produce only one half as much \rm H_2.

Alternatively, consider the ratio between the coefficient in front of \rm H_2 and \rm HCl is:

\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}.

\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol.

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as \rm 1.00\times 10^{5}\;Pa, that volume will be 22.7 liters.

V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L, or

V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L.

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

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