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matrenka [14]
3 years ago
8

Question 1:

Chemistry
1 answer:
Alona [7]3 years ago
5 0

Question 1:

(a) Sulfurous acid: H2SO3

Sulfuric acid: H2SO4

(b) Nitrous acid: H2NO2

Nitric acid: H2NO3

Question 2:

To calculate the pH, based on concentration of H+ ions, there is one formula:

ph =   - log_{10}(c(h + ))

So the pH of this solution is

-  log_{10}( {10}^{ - 9} )  = 9

(the solution is basic).

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DerKrebs [107]

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3 0
3 years ago
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What are examples of a Theory ?
galina1969 [7]

Paradoxes and Russell's Type Theories.

Simple Type Theory and the λ -Calculus.

Ramified Hierarchy and Impredicative Principles.

Type Theory/Set Theory.

Type Theory/Category Theory.

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6 0
3 years ago
Farmers who raise cotton once used arsenic acid, H3AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with
Bess [88]

Answer:

pH = 1.95

Explanation:

For polyprotic acids, it is generally assumed that all H⁺ comes from the 1st ionization step. The amount of H⁺ delivered into solution for the 2nd and 3rd ionization steps are in the order of 10⁻⁴M and 10⁻⁶M respectively and provide very little change in pH from the quantity delivered in the 1st ionization step.

Therefore... the [H⁺] concentraion and pH are computed as follows...

[H⁺] = √Ka₁[H₃AsO₄] = √(2.5 x 10⁻⁴)(0.500) M = 0.1118M

pH = -log[H⁺] = -log(0.01118) = 1.95

8 0
3 years ago
In the laboratory, a quantity of I2 was reacted with excess H2 to give 1.26 moles of HI. It is also known that the percent yield
Anon25 [30]

Answer:

1.008moles of iodine

Explanation:

Hello,

This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.

Percent yield = (actual yield / estimated yield) × 100

Actual yield = 1.2moles

Estimated yield = ?

Percentage yield = 84%

84 / 100 = 1.2 / x

Cross multiply and solve for x

100x = 84 × 1.2

100x = 100.8

x = 100.8/100

x = 1.008moles

1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI

5 0
3 years ago
Molar mass (NH4)2SO4
Hoochie [10]

Answer:

<u>132.15</u>

Explanation:

Molar mass N = 14.00

Molar mass H = 1.01

Molar mass H4 = 1.01 x 4 = 4.04

Molar mass NH4 = 14.00 + 4.04 = 18.04

Molar mass (NH4)2 = 18.04 x 2 = 36.08

Molar mass S = 32.07

Molar mass O = 16.00

Molar mass O4 = 16.00 x 4 = 64.00

Molar mass SO4 = 32.07 + 64.00 = 96.07

Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>

6 0
3 years ago
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