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Vlada [557]
2 years ago
5

(02.02 MC)

Chemistry
1 answer:
ratelena [41]2 years ago
7 0
This is due to evaporation
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What does it mean if two objects are in thermal equilibrium?
galina1969 [7]

Answer:

ang hirap nixbsxbaaaaksbs

6 0
2 years ago
Describe how you would prepare 350 ml of 0.100 m c12h22o11 starting with 3.00l of 1.50 m c12h22o11
Leona [35]

To prepare 350 mL of 0.100 M solution from a 1.50 M solution, we simply have to use the formula:

M1 V1 = M2 V2

So from the formula, we will know how much volume of the 1.50 M we actually need.

 

1.50 M * V1 = 0.100 M * 350 mL

V1 = 23.33 mL

 

So we need 23.33 mL of the 1.50 M solution. We dilute it with water to a volume of 350 mL. So water needed is:

350 mL – 23.33 mL = 326.67 mL water

 

 

Steps:

1. Take 23.33 mL of 1.50 M solution

<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M solution</span>

7 0
3 years ago
What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?
REY [17]

Answer:

CaN_{2} O_{6}

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is CaN_{2} O_{6}

5 0
2 years ago
List three ways the rate of solvation of sodium chloride in water may be<br> increased
r-ruslan [8.4K]

Answer:

1) Increasing temperature

2) Stirring

3) Increasing surface area  of salt by grinding it

5 0
3 years ago
A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib
DochEvi [55]

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

I_2+Br_2\rightarrow 2IBr

It is possible to set up the following equilibrium expression:

K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of x (reaction extent) would be:

0.0110=\frac{(2x)^2}{(0.50-x)^2}

Which can be solved for x to obtain two possible results:

x_1=-0.0277M\\\\x_2=0.0245M

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

[IBr]=2x=2*0.0249M=0.049M

Regards!

7 0
2 years ago
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