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snow_lady [41]
3 years ago
12

Please help with questions 14-17 ! and thanks - will mark you brainliest

Chemistry
1 answer:
KiRa [710]3 years ago
8 0

Answer: 14 :12 water molecules,  

Explanation: 17: 6 molecules

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What are standards or attributes of a design that can be measured.
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Explanation:

attribute of a person that often cannot be measured directly but can be assessed using numbers of indicators or manifest variables

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Three chemistry students measured the length of a copper bar. The recorded lengths were 5.05 cm, 5 cm , and 5.1 cm, What is the
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5.05 + 5 + 5.1 = 15.15cm Then you just divide it by the amount of measurements you had like this:15.15 ÷ 3 = 5.04999971cm Then you can just round it to the 3rd figure: 5.05cm < And that's the mean/average length of the bar. :) (Or the one above if you want all of the decimals too) 
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The molar mass of AI2CI6 is the mass of one mole of the compound. It is also which of the following?
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Is their a picture with the question?

Explanation:

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2 years ago
NEED ANSWER ASAP!<br> How many grams are in 1.50 moles of bromine liquid?
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119.85 grams Br or 120. grams Br (sig figs)

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1.50 moles Br          79.90 g Br

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7 0
3 years ago
Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra
maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4  mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

 

The equation for the solubilization reaction of Mg(OH)2 can be given as:

 

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

 Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²  

<u>Step 2:</u> Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

 X = <u>2.412 * 10^-4 mol/L = solubility in water</u>

<u>Step 3</u>: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+  = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X *(0.130)²  

5.61*10^-11 = X * (0.130)^2

X = <u>3.32*10^-9 = solubility in 0.130 M NaOH </u>

<u>Step 4:</u> Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0
3 years ago
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