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Aneli [31]
4 years ago
5

"solid potassium iodide decomposes into iodine gas and solid potassium. Write a a balanced chemical equation for this reaction"

Chemistry
1 answer:
wel4 years ago
3 0
KI (s)  ----\ \textgreater \     K(s)  + I_2 (g)

Now, balance the equation:

2KI (s)  ----\ \textgreater \     2K(s)  + I_2 (g)

I_2 in gaseous state exist as a diatomic molecule.
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Thermal energy travels<span> by conduction, convection, and radiation. It occurs when a cooler and warmer object touches each other. </span>
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3 years ago
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PLEASE HELP I DONT WANT TO FAIL!! ):
Fudgin [204]
Hope this helps :) I didn’t know how to write subscripts so I wrote it down on some paper.

3 0
3 years ago
Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.
Vsevolod [243]

The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of acid, H₃PO₄ (Ma) = 0.390 M
  • Volume of acid, H₃PO₄ (Va) = 24.5 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

Learn more about titration:

brainly.com/question/14356286

5 0
2 years ago
How many moles are in 54.0 grams of Silver?
Lelechka [254]
54g ag *(108mol ag/1 g ag) =5832mol ag
4 0
4 years ago
Small quantities of hydrogen can be prepared by the addition of hydrochloric acid to zinc. a sample of 195 ml of hydrogen was co
Vikentia [17]
Let's go over the given information. We have the volume, temperature and pressure. From the ideal gas equation, that's 4 out of 5 knowns. So, we actually don't need Pvap of water anymore. Assuming ideal gas, the solution is as follows:

PV=nRT
Solving for n,
n = PV/RT = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(25+273 K)
n = 7.897×10⁻³ mol H₂

The molar mass of H₂ is 2 g/mol.

Mass of H₂ = 7.897×10⁻³ mol * 2 g/mol = <em>0.016 g H₂</em>
5 0
3 years ago
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