Answer:
The answer to your question is below
Explanation:
Data
Volume 1 = V1 = 3000 l
Temperature 1 = T1 = 30°C
Pressure 1 = P1 = 99 kPa
Volume 2 = ?
Temperature 2 = T2 = 16°C
Pressure 2 = P2 = 45.5 kPa
Process
1.- To find the new volume we need to convert the temperature to °K
T1 = 30 + 273 = 303°K
T2 = 16 + 273 = 289°K
2.- Use the Combined gas law
P1V1/T1 = P2V2/T2
-Solve it for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (99 x 3000 x 289) / (303 x 45.5)
-Result
V2 = 85833000 / 13786.5
V2 = 6225.8 l
There are no conversion factor ratios.
Answer:
1026.88 kilocalories are released by the combustion of 1.50 mol of glucose
Explanation:
First we have to calculate the amount in grams in 1.5 moles of glucose by using formula
Molecular mass of glucose is 180.156 g/mol
Mass in grams = Molarity x Molecular mass
=1.50 moles x 180.156 g/mol
= 270.23 g
Now we will calculate the amount of energy released
Energy released by 1 gram of glucose = 3.8 Kcal
Energy released by 270.23 gram of glucose = 3.8 x 270.23 Kcal
= 1026.88 Kcal
1026.88 kilocalories are released by the combustion of 1.50 mol of glucose
A boulder sits at rest on top of a mountain. What conclusion can be made about the forces acting on the boulder? The forces acting on the boulder are balanced (net force equals zero).
Answer:
atoms bond mixes with oxygen
Explanation:
Just search it up on google dum dum
<h3>
Answer:</h3>
20 g
<h3>
Explanation:</h3>
We are given;
Original mass of K-40 before decay = 80 g
Half life of Potassium-40 = 1.25 billion years
Time taken by the decay = 2.5 billion years
We are required to calculate the mass of K-10 that will remain after 2.5 billion years.
Using the formula;
Remaining mass = Original mass × (1/2)^n
where n is the number of half lives
Number of half lives, n = Time taken ÷ half-life
= 2.5 billion years ÷ 1.25 billion years
= 2
Remaining mass = 80 g × (0.5)^2
= 80 g × 0.25
= 20 g
Therefore, the mass of K-40 isotope that will remain after 2.5 billion years is 20 g.