The volume of the new solution is
calculation
by use of the formula
M1V1= M2V2
M1 (molarity 1) = 2.13 M
V1 (volume 1) = 1.24 l
M2 ( molarity 2) = 1.60 M
V2 (volume 2) = ?
by making V2 the subject of the formula
V2 = M1 V1/ M2
V2=( 1.24 x 2.13)÷ 1.60= 1.651 L
Answer:
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Explanation:
Answer:
207.03°C
Explanation:
The following data were obtained from the question:
V1 (initial volume) = 6.80 L
T1 (initial temperature) = 52.0°C = 52 + 273 = 325K
P1 (initial pressure) = 1.05 atm
V2 (final volume) = 7.87 L
P2 (final pressure) = 1.34 atm
T2(final temperature) =?
Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:
P1V1/T1 = P2V2/T2
1.05 x 6.8/325 = 1.34 x 7.87/T2
Cross multiply to express in linear form as shown below:
1.05 x 6.8 x T2 = 325 x 1.34 x 7.87
Divide both side by 1.05 x 6.8
T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)
T2 = 480.03K
Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:
°C = K - 273
°C = 480.03 - 273
°C = 207.03°C
Therefore, the final temperature of the gas will be 207.03°C
Answer:
We'll have 82 moles ZnO and 41 moles S
Explanation:
Step 1: data given
Number of moles Zinc (Zn) = 82 moles
Number of moles sulfur oxide (SO2) = 42 moles
Step 2: The balanced equation
2Zn + SO2 → 2ZnO + S
Step 3: Calculate the limiting reactant
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
Zinc is the limiting reactant. It will completely be consume (82 moles). Sulfur oxide is in excess. There will react 82/2 = 41 moles
There will remain 42-41 = 1 mol SO2
Step 4: Calculate moles of products
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
For 82 moles Zinc we'll have 82 moles of Zinc Oxide (ZnO)
For 82 moles Zinc we'll have 82/2 = 41 moles of sulfur
We'll have 82 moles ZnO and 41 moles S
Answer:
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Explanation:
