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3 years ago

What is the minimum temperature needed to dissolve 35 grams of KCl in 100 grams of water?

Chemistry

1 answer:

egoroff_w [7]3 years ago

3 0

Answer:

$30^{\circ}\text{C}$

Explanation:

To know the temperature at which KCl dissolves in water we need to refer to the general solubility curves.

In the case of $KCl$ , $35\ \text{g}$ of it will dissolve in $100\ \text{g}$ of water at a minimum temperature of $30^{\circ}\text{C}$ .

So, the the minimum temperature needed to dissolve 35 grams of KCl in 100 grams of water is $30^{\circ}\text{C}$ .

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3 years ago

Arrange the following compounds in order of increasing acidity, and explain the reasons for your choice of order. Enter your ans

NNADVOKAT [17]

Answer:

(a) > (d) > (c) > (b)

Explanation:

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3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

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3 years ago

HELP HELP HELP HELPPP PLSSSSSS

Delvig [45]

Bestie since it’s Cl2, put 2 next to KCl. That means you need to 2 moles of K, so put 2 next to KI. That’s it

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