Answer:
a) Molarity KCl = 0.755 M
b) molality HNO = 5.09 m
Explanation:
- Formality (F) = moles sto / L sln
- Molarity (M) = # dissolved specie / L sln
- molality (m) = moles sto / Kg ste
- %p/p = ( g sto / g ste ) * 100
a) KCl ↔ K+ + Cl-
moles KCL:
⇒ 20 g KCl * ( mol / 74.6 g ) = 0.268 mol KCl
⇒ F = 0.268 mol KCl / 0.355 L = 0.755 F
⇒ M [ K+ ] = 1 * ( 0.755) = 0.755 M
b) 24% HNO:
calculation base: 1 g solution:
⇒24 = ( g sto / g sln) * 100
⇒ 0.24 = g sto / 1
⇒ g sto = 0.24g
⇒g sln = 1 - 0.24 = 076g sln
⇒Kg ste = 0.76 g * ( Kg / 1000g ) = 7.6 E-4 Kg ste
moles sto (HNO):
⇒ 0.24g * ( mol / 62.03g) = 3.869 E-3 moles HNO
⇒ m = 3.869 E-3 moles HNO / 7.6 E-4 Kg sln
⇒ m = 5.09 mol/Kg
Answer:
option D = N₂ + 3H₂ → 2NH₃
Explanation:
The reaction in which the nitrogen and hydrogen are reacted to form the ammonia is true.
N₂ + 3H₂ → 2NH₃
In this reaction we can see that one molecules of nitrogen react with three molecules of hydrogen and form two molecules of ammonia.
while the other options are in correct because,
2SO₂ + O₂ → 2SO₃
In this reaction two molecules of SO₂ react with one molecule of oxygen and form two molecules of SO₃.
2H₂S + 3O₂ → 2SO₂ + 2H₂O
In this reaction two molecules of H₂S react with three molecules of oxygen and produced four molecules of product.
SnO₂ + 2H₂ → Sn + 2H₂O
This is also incorrect because three molecules of reactant form three molecules of product.
Answer:
The correct option is: the precipitation of acidic compounds formed when components of air pollution interact with other components in the air.
Explanation:
<u>Acid rain</u> is a form of precipitation which is usually <u>acidic in nature</u>. The acidic nature of rain is due to the reaction of <u>water molecules with sulfur dioxide, carbon dioxide and nitrogen oxide present in the air, to produce acids.</u>
There are many adverse effects of acid rain, some of them are: effect on water bodies, forests, living organisms, corrosion of structures made of steel and weathering of buildings.
The balanced chemical equation is written as:
<span>CsF(s) + XeF6(s) ------> CsXeF7(s)
We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:
11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
