Answer:
5 × 10^-4 L
Explanation:
The equation of the reaction is;
2KClO3 = 2KCl + 3O2
Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles
From the stoichiometry of the reaction;
2 moles of KClO3 yields 3 moles of O2
0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas
From the ideal gas equation;
PV= nRT
P= 85.4 × 10^4 KPa
V=?
n= 0.165
R= 8.314 J K-1 mol-1
T= 40+273 = 313K
V= 0.165 ×8.134 × 313/85.4 × 10^4
V=429.4/85.4 × 10^4
V= 5 × 10^-4 L
values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6
location of the electron: In the 7th energy level away from the nucleus.
Explanation:
From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.
The orbital designation of the describe electron is 7d
- Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6
- The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus. Here it is 7.
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The answer to the question is D.
Answer:
36.66%
Explanation:
Step 1: Given data
- Mass of the sample: 2.875 g
Step 2: Calculate the mass of salt
The mass of the sample is equal to the sum of the masses of the components.
m(sample) = m(iron) + m(sand) + m(salt)
m(salt) = m(sample) - m(iron) - m(sand)
m(salt) = 2.875 g - 0.660 g - 1.161 g
m(salt) = 1.054 g
Step 3: Calculate the percent of salt in the sample
We will use the following expression.
%(salt) = m(salt) / m(sample) × 100%
%(salt) = 1.054 g / 2.875 g × 100% = 36.66%
Answer:
0.4 moles of water produced by 6.25 g of oxygen.
Explanation:
Given data:
Mass of oxygen = 6.25 g
Moles of water produced = ?
Solution:
Chemical equation;
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 6.35 g/ 32 g/mol
Number of moles = 0.2 mol
Now we will compare the moles of oxygen with water:
O₂ : H₂O
1 : 2
0.2 : 2×0.2 = 0.4 mol
0.4 moles of water produced by 6.25 g of oxygen.
