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ira [324]
3 years ago
10

H2(g)+p4(s)ph3(g) what the answer ?

Chemistry
1 answer:
disa [49]3 years ago
8 0

Answer:

6H2 + P4→ 4PH3

Explanation:

Phosphorus has 4 in it and hydrogen has 3 in it. in order to balance it, we have to put 4 in front of phosphine so that the phosphorus on the product side has an equal amount as to the one on the reactant side.

the only one left to balance is hydrogen and so in order to balance it we put a 6 on h2 because the hydrogen in the product size becomes 12 (4 * 3).

therefore the hydrogen on the reactant side becomes 12 as well (6 * 2)

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How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution
Lelechka [254]

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

  • The mole ratio of the acid, HCl (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
  • Volume of acid, HCl (Va) = 36 mL
  • Molarity of acid, HCl (Ma) = 1.63 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

Learn more about titration:

brainly.com/question/14356286

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6 0
2 years ago
What is the actual name of compound-s that is used to treat arthritis?
Rudiy27
Reichsteins substance if I remember correctly.

Hope this helped! :)

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3 years ago
A bag of potato chips contains 585 mL of air at 20.0 C and a pressure of 765 mmHg. Assuming the bag does not break, what will be
Scorpion4ik [409]

Answer:

volume = 972.23ml

Explanation:

using general gas law

P1V1/T1 = P2V2/T2

765 x 585/293 = 443 x V2/282

1527.39 =443 x V2/282

1527.38 x 282 = 443 x V2

430695.78 = 443 x V2

V2 = 430695.68/443

V2 = 972.23mL

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A force of attraction that holds atom together 
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3 0
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Electromagnetic radiation

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