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Lostsunrise [7]
3 years ago
10

How many atoms are in 55g of H2SO4? 3.4 atoms 2.58 atoms 04.14 atoms 98.1 atoms

Chemistry
2 answers:
pshichka [43]3 years ago
7 0

Answer: 3.4 atoms

Explanation: hope this helps you!!

Kamila [148]3 years ago
5 0
The answer to your question is 3.4 atoms
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Help needed ASAP, I will mark your answer as brainliest.
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I think the answer is C I think that bc I added 3 and the 2 which made it H5 and than multipled the 3 and the 2 which made it C6 than Cl
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A sample of Br 2 ( g ) Br2(g) takes 42.0 min 42.0 min to effuse through a membrane. How long would it take the same number of mo
Simora [160]

15 mins

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6 0
3 years ago
_____ compounds include a prefix on the first and second elements and a suffix on the second element, whereas _____ compounds do
Anit [1.1K]

Answer:

Molecular compound and ionic compound

Explanation:

Molecular compounds include a prefix on the first and second elements and a  suffix on the second element, whereas Ionic compounds do not have prefixes and only include a suffix on the second element.

4 0
3 years ago
How is the miscibility of two liquids related to their polarity?
makvit [3.9K]

Answer:

The more polar the liquid, the more likely that it is miscible with water. The polarity of a liquid does not affect its miscibility with water. The less polar the liquid, the more likely that it is miscible with water. The more polar the liquid, the less likely that it is miscible with water.

Explanation:

hope it helps you

4 0
3 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
3 years ago
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