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wlad13 [49]
3 years ago
13

HClO4 acid solution has a concentration of 5 Molarity. Calculate the concentration of this solution in

Chemistry
1 answer:
Nutka1998 [239]3 years ago
6 0

Answer:

1. Percentage by weight = 0.5023 = 50.23 %

2. molar fraction =0.153

Explanation:

We know that

Molar mass of HClO4 = 100.46 g/mol

So the mass of 5 Moles= 5 x 100.46

       Mass (m)= 5 x 100.46 = 502.3 g

Lets assume that aqueous solution of HClO4  and the density of solution is equal to density of water.

Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.

We know that

Mass = density x volume

Mass of 1000 ml  solution = 1 x 1000 =1000     ( density = 1 gm/ml)

            m'=1000 g

1.

Percentage by weight = 502.3 /1000

Percentage by weight = 0.5023 = 50.23 %

2.

We know that

molar mass of water = 18 g/mol

mass of water in 1000 ml = 1000 - 502.3 g=497.9 g

So moles of water = 497.7 /18 mole

moles of water = 27.65 moles

So molar fraction = 5/(5+27.65)

molar fraction =0.153

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Answer:

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8 0
2 years ago
Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


4 0
3 years ago
Why is oxygen written with a subscript 2 in CO2?
Alexeev081 [22]

Answer:

Explanation:

the answer is b

6 0
3 years ago
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