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mestny [16]
2 years ago
8

nter your answer in the provided box. Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. C

alculate the pH of a HNO3 solution having a hydrogen ion concentration of 0.39 M. pH
Chemistry
1 answer:
OLga [1]2 years ago
8 0

Answer:

Explanation:

pH = - log[ H⁺ ]

[ H⁺ ] = .39

pH = - log .39

= - log (39 x 10⁻²)

= 2 - log 39

= 2 - 1.59

=  .4 1

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What does this picture represent?
WARRIOR [948]

the answer is d electronic balance

8 0
2 years ago
Question 6 (3 points)
olga55 [171]

Answer:

Explanation:

Given parameters:

Molarity of KOH = 0.26M

Volume of H₂SO₄ = 19.76mL

Molarity of H₂SO₄ = 0.20M

Unknown:

Volume of KOH = ?

Solution:

This is a neutralization reaction in which an acid reacts with a base to produce salt and water:

         H₂SO₄     +     2KOH      →    K₂SO₄    +     2H₂O

 We solve from the known to the unknown in the reaction.

The known is the acid and from there we can find the number of moles of the acid to be completely neutralized:

       Number of moles of acid = molarity x volume

       Number of moles of acid = 19.76 x 0.20 = 3.95mol

From the balanced reaction equation:

      1 mole of acid reacts with 2 moles of the bases KOH

     3.95mole of acid would react with 3.95moles x 2 of the base

  Number of moles of reacting base = 7.90moles

To find the volume of base;

  Volume of base = \frac{number of moles of base}{molarity}

   Volume = \frac{7.90}{0.26} = 30.40mL

Learn more:

Neutralization brainly.com/question/6447527

#learnwithBrainly

8 0
3 years ago
Pls answer this question and have a AMAGING day!! :))
ladessa [460]

Answer:

B

Explanation:

because it is being cooled down

hoped this helps

3 0
2 years ago
15 POINTS PLEASE HELP
Lapatulllka [165]

Answer:

Neutrons = ( Atomic mass – Atomic number ) ( A–Z )

Protons and Electrons are equal to the atomic number

For example Neon,

Mass number (A) = 20

Atomic Number (Z) = 10

Number of Protons = 10

Number of Electrons = 10

Number of Neutrons = ( A–Z ) = 10

Electronic distribution :

K= 2

L= 8

4 0
3 years ago
Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and
andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}

 moles of N_{2} + moles of H_{2} = 0.908 mol

Since, 2 moles of N_{2} = 2 \times 0.227 = 0.454 mol

As it is known that the ideal gas equation  is PV = nRT

And, as the temperature and volume were kept constant, so we can write

        \frac{P(_in)}{n_(in)} = \frac{P_(final)}{n_(final)}

          \frac{10.4}{0.908} = \frac{P_(final)}{0.454
}

       P_(final) = 10.4 \times \frac{0.454}{0.908}

                            = 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0
2 years ago
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