Answer:
8.68 L is the new volume
Explanation:
You use Boyle's law for this.
= first pressure
= second pressure
= first volume
= second volume
Convert pressure from atm to mmHg (use same units):
5.97 x 760 = 4537.2 -> 4.54 x 10³
...maintain 3 significant figures in calculation, and round as needed...
(4.54 x 10³ mmHg)(2.79 L) = (1460 mmHg)()
(4.54 x 10³ mmHg)(2.79 L) / (1460 mmHg) = = 8.68 L
Hope this helps :)
<u>Given:</u>
Mass of solvent water = 4.50 kg
Freezing point of the solution = -11 C
Freezing point depression constant = 1.86 C/m
<u>To determine:</u>
Moles of methanol to be added
<u>Explanation:</u>
The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:
ΔTf = kf*m
where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution
ΔTf = 0 C - (-11.0 C) = 11.0 C
m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg
11.0 = 1.86 * moles of methanol/4.50
moles of methanol = 26.613 moles
Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.
Decay is a type of degradation reaction and thus is considered a first order reaction. thus the formula goes like this.
rate constant= 0.693/half life
so here...
rate constant= 0.693/1620 year^-1
Different elements produce different colors of light when heated because the electrons in these elements have different permissible energy levels. When an element is heated, the electrons inside it become excited and move to an higher energy level from the ground state. When the electrons drop from this higher energy level, they typically emit energy quantum, the color of the light that is observed at this stage depends on difference that exist in the two energy levels.<span />
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
