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kow [346]
3 years ago
14

Accurately calculate the concentration of the diluted solution starting with the concentration of the stock fe(no3)3 solution (0

.2 m).
Chemistry
1 answer:
elena-s [515]3 years ago
8 0
Ans: 25 mM

For example, suppose we take the following concentrations:

Diluted Fe(NO_{3})_{3} =1 ml 

<span>1 M KSCN = 5 ml 
</span>
0.1 M HNO_{3}<span> = 2  ml 
</span>----------------------------------------
<span>Then, total volume = 8 ml 

Now, 
</span> <span>final concentration of  diluted Fe</span>(NO_{3})_{3}<span> =</span>\frac{Concentration - of Fe(NO_{3})_{3}*Volume of Fe(NO_{3})_{3}}{Total Volume} \\ \\<span> 
</span>
given stock solution is 0.2 M

Thus, concentration of diluted solution = \frac{0.2*1}{8} = 0.025 M = 25 milli M
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erastovalidia [21]

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Explanation:

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8 0
3 years ago
Consider this reaction:
nadya68 [22]
11.7 g hope this helps and have a great day
4 0
2 years ago
Does the reaction between sodium
Masteriza [31]

The reaction between sodium

and chlorine is NaCl which is physical change

8 0
4 years ago
1 pts
soldier1979 [14.2K]

Answer:

144 g

Explanation:

Use the mole ratio of 4 mol CO2 for every 9 mol O2 to convert from mol O2 to mol CO2. Then use the molar mass of CO2 to convert from mol of CO2 to grams of CO2.

7.34 mol O2 • (4 mol CO2 / 9 mol O2) • (44.01 g CO2 / 1 mol CO2) = 144 g CO2

5 0
3 years ago
Read 2 more answers
2. Consider the reaction 2NO(g) + O2(g) → 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is r
vredina [299]

Answer :

(a) The rate of NO_2 formed is, 0.066 M/s

(b) The rate of O_2 formed is, 0.033 M/s

Explanation : Given,

\frac{d[NO]}{dt} = 0.066 M/s

The balanced chemical reaction is,

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate of disappearance of NO = -\frac{1}{2}\frac{d[NO]}{dt}

The rate of disappearance of O_2 = -\frac{d[O_2]}{dt}

The rate of formation of NO_2 = \frac{1}{2}\frac{d[NO_2]}{dt}

As we know that,

\frac{d[NO]}{dt} = 0.066 M/s

(a) Now we have to determine the rate of NO_2 formed.

\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}

\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s

The rate of NO_2 formed is, 0.066 M/s

(b) Now we have to determine the rate of molecular oxygen reacting.

-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}

\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s

The rate of O_2 formed is, 0.033 M/s

6 0
3 years ago
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