Answer:- 14.0 moles of hydrogen present in 2.00 moles of ![[tex]C_6H_7N](https://tex.z-dn.net/?f=%20%5Btex%5DC_6H_7N)
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Solution:- We have been given with 2.00 moles of 
and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:
= 14.0 g H
So, there are 14.0 g of hydrogen in 2.00 moles of .
Answer:
3.51× 10²³ formula units
Explanation:
Given data:
Mass of CaO = 32.7 g
Number of formula units = ?
Solution:
First of all we will calculate the number of moles.
Number of moles = mass/molar mass
Number of moles = 32.7 g/ 56.1 g/mol
Number of moles = 0.583 mol
Number of formula units:
1 mole = 6.022 × 10²³ formula units
0.583 mol × 6.022 × 10²³ formula units / 1 mol
3.51× 10²³ formula units
The number 6.022 × 10²³ is called Avogadro number.
Answer: 20 mg Te-99 remains after 12 hours.
Explanation: N(t) = N(0)*(1/2)^(t/t1/2)
N(t) = (80 mg)*(0.5)^(12/6)
N(t) = 20 mg remains after 12 hours
Answer:
Digestion of food.
Explanation:
I hope my answer help you.
Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
The work:
If it is an ideal gas:
Solving:
Replacing:
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
