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Leokris [45]
3 years ago
13

What is the [SO32-) in the 0.080 M H2SO3 .080 M Solution?

Chemistry
1 answer:
slega [8]3 years ago
4 0

Answer:

0.080 M

Explanation:

From the question given above, the following data were obtained:

Concentration of H₂SO₃ = 0.080 M

Concentration of SO₃²¯ =…?

The concentration of SO₃²¯, can be obtained as follow:

H₂SO₃ (aq) <==> 2H⁺ (aq) + SO₃²¯ (aq)

From the balanced equation for above,

1 mole of H₂SO₃ produced 1 mole of SO₃²¯.

Therefore, 0.080 M H₂SO₃ will also produce 0.080 M SO₃²¯.

Thus, the concentration of SO₃²¯ is 0.080 M

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The scientific evidence that scientists use in supporting or critiquing the conclusions of experiments usually consists observations based on large sample sizes. 
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3 years ago
A runner averages 8 minutes and 25 seconds per mile. What is her average velocity in miles per hour?
labwork [276]

8 minutes and 25 seconds is the same as 8.25 in decimal form.

Divide by 60 minutes.

60 / 8.25 = 7.27 miles per hour

Best of Luck!

6 0
3 years ago
How many valence electrons are there in sulfur, and iodine.
Vinil7 [7]

Today I will explain what a valence electron is and how many valence electrons each of the atoms your asking about have. :)

→ A valence electron is an electron in the outer shell of an atom that is able to participate in bonding. While iodine has 53 electrons, all but seven are in the inner shells of the iodine atom.

→ Sulfur has sixteen electrons in total, two in the inner shell, eight in the middle, and six in the outer. This means that sulfur has six valence electrons.

↑   ↑   ↑   Hope this helps! :D

7 0
3 years ago
Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2
Karolina [17]

Answer:

The mole fractions:

x_{HBr}=\frac{100mol}{318.5}=0.314

x_{Br_2}=\frac{78mol}{318.5}=0.245

x_{H_2O}=\frac{78mol}{318.5}=0.245

x_{O_2}=\frac{62.5mol}{318.5}=0.196

Explanation:

The reaction described is:

2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78

(the 0.78 is because of the fractional conversion)

n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25

(the 1.25 is because of the oxygen excess)

n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

nn_{HBr}=100 mol HBr

n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *100mol HBr*0.78

n_{Br_2}=78 mol Br2

n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *100 mol HBr*1.25

n_{O_2}=62.5 mol O_2

n_{H_2O}=n_{Br_2}= 78 mol

n_{total}=(78+78+100+62.5)mol= 318.5mol

The mole fractions:

x_{HBr}=\frac{100mol}{318.5}=0.314

x_{Br_2}=\frac{78mol}{318.5}=0.245

x_{H_2O}=\frac{78mol}{318.5}=0.245

x_{O_2}=\frac{62.5mol}{318.5}=0.196

4 0
3 years ago
What is the formula of haso4​
ladessa [460]

Explanation:

The structure of Ferrarrisite Ca5(HAs O4)2(AsO4)2

8 0
3 years ago
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