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Igoryamba
3 years ago
15

Can 750 mL of water dissolve 0.60 mol of gold(III) chloride (AuCl3)?

Chemistry
1 answer:
nydimaria [60]3 years ago
3 0

Answer:

Yes  

Explanation:

1. Mass of 0.60 mol of AuCl₃  

\text{Mass} = \text{0.60 mol} \times \dfrac{\text{303.33 g}}{\text{1 mol}} = \text{184 g}

2. Mass of AuCl₃ in 750 mL

The solubility of AuCl₃ is 68 g/100 mL.

In 750 mL of water, you can dissolve

\text{Mass of AuCl}_{3} = \text{750 mL} \times \dfrac{\text{68 g}}{\text{100 mL}} = \text{510 g AlCl}_{3}

∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.

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if the percent by volume is 2% and the volume of solution is 250 mL what is the volume of solute and solution
Liula [17]

Answer:

Solute = 5 mL; solution = 250 mL  

Explanation:

The formula for percent by volume is

\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 2 % v/v,

\begin{array}{rcl}2 \, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\2 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{2 \times 250\text{ mL}}{100}\\\\ & = & \textbf{5 mL}\\\end{array}

If there is no change of volume on mixing,

Volume of solution = 250 mL

 -Volume of solute = <u>     </u><u>5</u><u>      </u>

 Volume of solvent = 245 mL

5 0
3 years ago
Why are hydrogen, lithium, and sodium classified as reactive elements?
Sergeu [11.5K]
<span>The alkali metals and hydrogen are reactive because they have only one electron to give in order to complete their valence shell. It is easier to give that one electron so when given the opportunity they will. This means they will react with anything polar or willing to take an electron.</span>
4 0
4 years ago
Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.
Dovator [93]
Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10  5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2  2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

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