22.0 is the same as saying that in 100 grams of a chocolate bar, there are 22.0 grams of pecans. or to make it easier because of this problem- 100 Kilograms of a chocolate bar, there is 22.0 Kg of pecans. we can use this as a conversion factor (what is used to convert a value to another value.
conversion factor---> 22.0 kg of pecan= 100 kg of chocolate bar
Note: remember this, what you are converting from goes in the denominator, what you converting to goes in the numerator.
5.0 Kg of pecan (100 Kg of chocolate bar/ 22.0 Kg of pecan)= 23 Kg of chocolate bar
Answer:
Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]
Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]
Explanation:
An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:
HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)
And Ka, the acid dissociation constant is:
<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />
When base:
HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)
And kb, base dissociation constant is:
<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>
Answer:
See explanation
Explanation:
A flammable solvent refers to a solvent that catches fire easily. The precautions to be taken when working with flammable solvents are;
1) heat the solvent at a low to medium hot plate setting.
2) if you need to boil the solvent, use a condenser rather than a flask or beaker without a cover.
3) make sure that the hotplate is larger than the vessel containing the mixture that is being heated.
4) do not use strong oxidizing agents
Answer:
0.5
Explanation:
2NaCl(s) + 2H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2H2O(g) + Cl2(g)
Using ideal gas equation,
PV = nRT
28.7torr
Converting torr to atm,
= 0.0378atm
V = 0.597L
T = 27 °C
= 300 K
a) PV = nRT
(0.0378atm) * (0.597L) = n(0.0821) * (300k)
= 0.000915 mol
moles of water and chlorine = 0.000915 mol
From the above equation, the ratio of water to chlorine = 1 : 2
Therefore, mole of chlorine = 0.000915/2
= 0.000458 mol
mole fraction = moles of specie/moles of all the species present
= 0.000458/0.000915
= 0.5
The only compound that contains covalent bonds would be A. BCl4-.
