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KiRa [710]
3 years ago
13

A student pipets exactly 5.00ml of 3.47 x 10.2M FeCl3 solution into a Vol flask and adds enough water to it to make 250ml of sol

ution. What is the concentration of the diluted solution?
Chemistry
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

The concentration would be 0.76 mol/L.

The most common way to solve this problem is to use the formula

c1V1=c2V2

In your problem,

c1 = 4.2 mol/L; V1 = 45.0 mL

c2 = ?; V2 = 250 mL

c2=c1×V1V2 = 4.2 mol/L × 45.0mL250mL = 0.76 mol/L

This makes sense. You are increasing the volume by a factor of about 6, so the concentration should be about ¹/₆ of the original (¹/₆ × 4.2 = 0.7).

Explanation:

if it is helpful...... plzz like and follow

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Why is the ideal gas law useful although ideal gases do not exist
babymother [125]
I think it would be useful because then a gas station would have to save gas, instead of selling it all. But, for this question Im not to sure. Is this a multiple choice question_ 

8 0
3 years ago
Mass of moist aggregate sample as brought to the laboratory: 5,306 grams Mass of oven dried aggregate: 5,229 grams Mass of aggre
Lerok [7]

Answer:

a.   2.748

b. 68.91

c. 1.45%

d. 37.84%

Explanation:

To determine the:

a. The aggregate's bulk Saturated Surface Dry specific gravity

Given:

The mass of the moist aggregate sample is 5,229 g.

The mass of the oven-dried aggregate is 5,306 g.

The mass of the aggregate submerged in water is 5,229 g.

The mass of the Saturated Surface Dry aggregate is 3,298 g.

The aggregate's bulk SSD specific gravity =the dry weight ÷ ( the SSD weight - the submerged weight)

Given:

The mass of the oven-dried aggregate is 5,306 g.

The mass of the Saturated Surface Dry aggregate is 3,298 g.

The mass of the aggregate submerged in water is 5,229 g.

= 5,306 ÷ ( 3,298 - 5,229)

= 5,306 ÷ 1,931 = 2.748

b. The aggregate appearance specific gravity

the aggregate apparent specific gravity = the dry weight ÷ (the dry weight - the submerged weight)

Given:

The mass of the oven-dried aggregate is 5,306 g.

The mass of the aggregate submerged in water is 5,229 g

The aggregate apparent specific gravity = 5,306  ÷ (5,306 - 5,229)

The aggregate apparent specific gravity = 68.91

c. The moist content of stockpile aggregate

The moisture content of stockpile aggregate = ( the moist aggregate weight - the dry weight) ÷ the dry weight

Given:

The mass of the moist aggregate sample is 5,229 g.

The mass of the oven-dried aggregate is 5,306 g

The moisture content of stockpile aggregate = (5,229 - 5,306)  ÷ 5,306

The moisture content of stockpile aggregate = 0.0147 x 100% (when expressed in percentage) = 1.45%

d. Absorption

Absorption =  (mass of the Saturated Surface Dry aggregate - mass of the oven-dried aggregate)  ÷ mass of the oven-dried aggregate x 100%

Given:

The mass of the Saturated Surface Dry aggregate is 3,298 g

The mass of the oven-dried aggregate is 5,306 g.

Absorption =  (3,298  - 5,306)  ÷ 5,306  x 100% = 37.84%

8 0
3 years ago
1.72 moles of NOCI were placed in a 2.50 L reaction chamber
elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI  remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

   2NOCI(g) = 2NO(g) + Cl2(g).

I    1.72

C   0.56           0.56         0.28

E   1.16              0.56         0.28

Molarity at equilibrium :

NOCl :

\tt \dfrac{1.16}{2.5}=0.464

NO :

\tt \dfrac{0.56}{2.5}=0.224

Cl2 :

\tt \dfrac{0.28}{2.5}=0.112

\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026

4 0
3 years ago
8. What is the name of this compound?
olga2289 [7]

Answer:

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Explanation:

sec-butyl bromide

CH3 -CH-CH2-CH3

         Br

tert-butyl bromide

       CH3

CH3-C- Br

       CH3

isopropyl bromide

       CH3

CH3-CH-Br

isobutyl bromide

       CH3

CH3-CH-CH2-Br

6 0
4 years ago
Find the numberof moles in 60 grams of SO3
ziro4ka [17]

The answer is in the photo

8 0
3 years ago
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