<h3>
Answer:</h3>
0.248 g
<h3>
Explanation:</h3>
<u>We are given;</u>
- Time as 40 minutes
- Original amount of N-13
We are required to calculate the amount left after decay.
- We need to know the half life of N-13 first;
- Half life of N-13 is 9 minutes 58 seconds (9.965 minutes0
We are going to use the formula;
- Remaining amount = Original amount × 0.5^n
That is; N = N₀ × 0.5^n , where n is the number of half lives.
n = 40 minutes ÷ 9.965 minutes
= 4.014
Therefore;
N = 4 g × 0.5^4.014
= 0.248 g
Therefore, the mass of Nitrogen-13 that will be left is 0.248 g
Answer:
a) 2.6 mole of O2 per mole of methane
b) 9.78 mole of N2 per mole of methane
c) Mole fraction of H2O = 0.33
d) 6.12 mole of H2O pero mole of methane
Explanation:
The first thing you have to do is the balanced reaction of the methane combustion:
CH4 +2 O2 -->2 H2O + CO2
You calculate the moles of O2 needed to react with 1 mole of methane and add the excess factor
2 x 1.3 = 2.6 moles of O2
For the N2 moles you calculate it with the air composition (21 % O2, 79% N2)
2.6 moles of O2 * 0.79/0.21 = 9.78 moles of N2
With the total stream of air = 12.38 moles you add the humidity factor
12.38 * 1.5 = 18.57 moles So 6.12 are moles of H20 entering per mole of CH4.
To calculate the mole fraction yo divide the moles of water among the moles of the stream:
6.12/18.57 --> 0.33
Answer:
Volume in L = 0.028 L
Explanation:
Given data:
Mass of copper phosphate = 27.6 g
Molarity solution = 6.03 M
Volume of solution = ?
Solution:
First of all we will calculate the number of moles of copper phosphate.
Number of moles = mass/molar mass
Number of moles = 27.6 g/ 159.61 g/mol
Number of moles = 0.17 mol
Volume of solution:
Molarity = number of moles / volume in L
By putting values,
6.03 M = 0.17 mol / volume in L
Volume in L = 0.17 mol / 6.03 M
Volume in L = 0.028 L
Answer:
The answer to your question is: T = 546°K
Explanation:
V1 = 100 l
T1 = 0°C = 273°K
V2 = 200 l
T2 = ?
Pressure = constant
Formula
Charles law relates volume vs temperature
Substitution
Result
T2 = 546°K
