Answer:
It is <em><u>photosynthesis</u></em><em><u> </u></em><em><u>and </u></em><em><u>transpiration</u></em><em><u> </u></em>
hope it helps
Answer:
0.03atm
Explanation:
Given parameters:
Total pressure = 780torr
Partial pressure of water vapor = 1.0atm
Unknown:
Partial pressure of radon = ?
Solution:
A sound knowledge of Dalton's law of partial pressure will help solve this problem.
The law states that "the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".
Mathematically;
P
= P
+ P
+ P
Since the total pressure is 780torr, convert this to atm;
760torr = 1 atm
780torr =
atm = 1.03atm
For this problem;
Total pressure = Partial pressure of radon + Partial pressure of water vapor
1.03 = Partial pressure of radon + 1.0
Partial pressure of radon = 1.03 - 1.00 = 0.03atm
78.6 g (1mol/60.1 g)= 1.31 moles of isopropanol
Answer: 131 g of bromine is required.
Explanation:
The balanced equation will be :

To calculate the moles, we use the equation:
moles of
According to stoichiometry :
2 moles of
require = 3 moles of
Thus 0.544 moles of
require=
of
Mass of
Thus 131 g of bromine is required.
Answer:
7.04 g
Explanation:
Let's consider the reaction in the last step of the Ostwald process.
3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)
The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 6.40 g are:
6.40 g × (1 mol/63.01 g) = 0.102 mol
The molar ratio of NO₂ to HNO₃ is 3:2. The reacting moles of NO₂ are:
0.102 mol HNO₃ × (3 mol NO₂/2 mol HNO₃) = 0.153 mol NO₂
The molar mass of NO₂ is 46.01 g/mol. The mass corresponding to 0.153 moles is:
0.153 mol × (46.01 g/mol) = 7.04 g