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3 years ago

What is the concentration of a solution formed by mixing 89.94 grams of cobalt (II)

Chemistry

2 answers:

gulaghasi [49]3 years ago

6 0

Answer:

$M=0.467M$

Explanation:

Hello,

In this case, cobalt (II) molar mass is 280.996 g/mol since it is about heptahydrate, so we compute the moles:

$n=89.94g*\frac{1mol}{280.996g}= 0.32mol$

Then, since we need the volume in litres only:

$V=68.6cL*\frac{1L}{100cL} =0.686L$

Finally, we compute the molarity:

$M=\frac{0.32mol}{0.686L}\\ \\M=0.467M$

So your answer should be refined.

Best regards.

bezimeni [28]3 years ago

3 0

Answer:

0.47 M

Explanation:

The concentration of the solution can be calculated using the following equation:

$C = \frac{\eta}{V} = \frac{m}{M*V}$

<u>Where:</u>

V: is the volume of the solution = 68.6x10⁻² L

η: is the moles of cobalt (II) sulfate

m: is the mass of cobalt (II) sulfate = 89.94 g

M: is the molar mass of cobalt (II) sulfate = 281.103 g/mol

The concentration of cobalt (II) sulfate is:

$C = \frac{m}{M*V} = \frac{89.94 g}{281.103 g/mol*68.6 \cdot 10^{-2} L} = 0.47 mol/L$

We used the molar mass of the cobalt (II) sulfate heptahydrate (281.103 g/mol) since it is one of the most common salts of cobalt.

Therefore, the concentration of a solution of cobalt (II) sulfate is 0.47 M (assuming that the cobalt (II) sulfate is heptahydrate).

I hope it helps you!

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2 years ago

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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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If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?

Nata [24]

Answer:

The correct answer is option B.

Explanation:

$3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2$