We should describe a little bit the legend.
A - Element - we should have circles with same color and not bonded together (argon gas).
B - Compound - here we may have circles with same or different color bonded together (water or oxygen which is a diatomic molecule).
C - Mixture of elements - circles with different colors not not bonded together (mixture of noble gases).
D - Mixture of compounds - circles with same or different color bonded together but we should see two or more types of connectivity between circles (mixture of water and ethanol).
E - Mixture of elements and compounds - circles with same or different color bonded together mixed with circles with same color and not bonded together (a mixture between oxygen which is a diatomic molecule and noble gas like argon).
Now we may answer the question:
1) B
2) C
3) D
4) D
5) A
6) B
7) B
8) E
9) E
10) D
11) B
12) D
13) D
14) D
15) D
Answer:
ghrtdfghtyfrhjbfyrt
Explanation:
t6n yrtnyrt tryrbhyrty ryrhtr
Answer: Yes, your answer is correct.
The question is incomplete, here is the complete question:
Calculate the mole fraction of the ionic species KCl in the solution A solution was prepared by dissolving 43.0 g of KCl in 225 g of water.
<u>Answer:</u> The mole fraction of KCl in the solution is 0.044
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of water = 225 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:
Given mass of KCl = 43 g
Molar mass of KCl = 74.55 g/mol
Putting values in equation 1, we get:
Mole fraction of a substance is given by:
Moles of KCl = 0.577 moles
Total moles = [0.577 + 12.5] = 13.077 moles
Putting values in above equation, we get:
Hence, the mole fraction of KCl in the solution is 0.044
Answer:
2H2S + 3O2 → 2SO2 + 2H2O
Explanation:
Step 1: Data given
Hydrogen sulfide = H2S
Oxygen = O2
sulfur dioxide = SO2
water = H2O
Step 2: The unbalanced equation
H2S + O2 → SO2 + H2O
Step 3: Balancing the equation
H2S + O2 → SO2 + H2O
On the left side we have 2x O (in O2) and on the right side we have 3x O (2x in SO2 and 1x in H2O). To balance the amount of O, we have to multiply O2 (on the left side) by 3 and SO2 and H2O on the right side by 3.
H2S + 3O2 → 2SO2 + 2H2O
On the right side we have 4x H and on the left side we have 2x H. To balance the amount of H, we have to multiply H2S by 2.
Now the equation is balanced.
2H2S + 3O2 → 2SO2 + 2H2O
