Which of the processes of the water cycle occur by releasing energy
1 answer:
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Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
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In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:
2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:
3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:
4. Here, since the molar mass of ammonia is 17.03 g/mol:
5. Here, since the molar mass of SO2 is 64.06 g/mol:
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There is a specific formula to use for these type of problems.
ln (P2/ P1)= Δvap/ R x (1/T1 - 1/T2)
R= 8.314
P1= 92.0 torr
T1= 23 C + 273= 296 K
P2= 351.0 torr
T2= 45.0 C + 273= 318 K
plug the values and solve for the unknown
ln( 351.0/ 92.0)= Δvap/ 8.314 x (1/296 - 1/318)
Δvap= 47630.6 joules
ln (P2/ P1)= Δvap/ R x (1/T1 - 1/T2)
R= 8.314
P1= 92.0 torr
T1= 23 C + 273= 296 K
P2= 351.0 torr
T2= 45.0 C + 273= 318 K
plug the values and solve for the unknown
ln( 351.0/ 92.0)= Δvap/ 8.314 x (1/296 - 1/318)
Δvap= 47630.6 joules
Answer:
2578.99 years
Explanation:
Given that:
100 g of the wood is emitting 1120 β-particles per minute
Also,
1 g of the wood is emitting 11.20 β-particles per minute
Given, Decay rate = 15.3 % per minute per gram
So,
Concentration left can be calculated as:-
C left =
Where,
is the concentration at time t
is the initial concentration
Also, Half life of carbon-14 = 5730 years
Where, k is rate constant
So,
The rate constant, k = 0.000120968 year⁻¹
Time =?
Using integrated rate law for first order kinetics as:
So,
<u>t = 2578.99 years</u>