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sertanlavr [38]
3 years ago
5

100 pointts pls answer quick and no link :In the reaction, Hydrogen + Iodine --> Hydrogen Iodide at equilibrium, some Iodine

is is added. What happens to the equilibrium?
Chemistry
1 answer:
sesenic [268]3 years ago
4 0

Answer:

the concentration of hydrogen iodide will be higher than it was in the original equilibrium conditions.

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What is the molarity of a solution of h3po4 if 10.2 ml is neutralized by 53.5 ml of 0.20m koh what is the molarity of a solution
Anna35 [415]

The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M

<h3>Balanced equation </h3>

H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 1
  • The mole ratio of the base, KOH (nB) = 3

<h3>How to determine the molarity of H₃PO₄ </h3>
  • Volume of acid, H₃PO₄ (Va) = 10.2 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
  • Volume of base, Ca(OH)₂ (Vb) = 53.5 mL
  • Molarity of acid, H₃PO₄ (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 10.2) / (0.2 × 53.5) = 1 / 3

(Ma × 10.2) / 10.7 = 1 / 3

Cross multiply

Ma × 10.2 × 3 = 10.7

Ma × 30.6 = 10.7

Divide both side by 30.6

Ma = 10.7 / 30.6

Ma = 0.35 M

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8 0
2 years ago
An ionic bond is between what kind of charged particles?
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Answer: An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between atoms.

Explanation:

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Latoya performs an experiment using the lab equipment shown. Which two properties is Latoya most likely measuring?
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Answer: Volume and temperature

Explanation:

5 0
4 years ago
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The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab
mixas84 [53]

Answer:

See the answers below

Explanation:

1)  100. mL of solution containing 19.5 g of NaCl  (3.3M)

2)  100. mL of 3.00 M NaCl solution (3 M)

3) 150. mL of solution containing 19.5 g of NaCl  (2.2 M)

4)  Number 1 and 5 have the same concentration (1.5M)

MW of NaCl = 23 + 36 = 59 g

For number 3

          59 g ------------------- 1 mol

           19,5 g -----------------   x

  x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For number 4,

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For number 5

Molarity (M) = 0.450/0.3 = 1.5 M

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