Answer:
The new temperature will be 2546 K or 2273 °C
Explanation:
Step 1: Data given
The initial temperature = 1000 °C =1273 K
The volume = 20L
The volume increases to 40 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 20L
⇒with T1 = the initial temperature = 1273 K
⇒with V2 = the increased volume = 40L
⇒with T2 = the new temperature = TO BE DETERMINED
20L/ 1273 K = 40L / T2
T2 = 40L / (20L/1273K)
T2 = 2546 K
The new temperature will be 2546 K
This is 2546-273 = 2273 °C
Since the volume is doubled, the temperature is doubled as well
Answer:
[CH₃COOH] = 17.4 M
Explanation:
Acetic acid → CH₃COOH
Molar mass → 60 g/mol
99.5% is percent by mass concentration. It means that 99.5 grams of solute are contained in 100g of solution.
Density → 1.05 g/mL. This data is always referred to solution, not solute!.
We determine solution's volume:
1.05 g/mL = 100 g / V → V = 100 g /1.05 g/mL → 95.2 mL
Now we know, that 99.5 g of acetic acid are contained in 95.2 mL
Let's convert to mmoles → 99.5 g / 60 g/mol = 1.66 moles
We convert solution's volume to L → 95.2 mL . 1L / 1000 mL = 0.0952 L
M (mol/L) = 1.66 mol / 0.0952 L = 17.4 M
Answer:
Because it lose detectable mass.
Explanation:
Law of conservation of mass states that substances involved in chemical reaction do not lose or gain any mass. So the above reactions don't demonstrate the law of conservation of mass because it lose mass
Answer:
we needa picture luv we don't know the words are
Answer:
Number of moles of oxygen used=13.5moles
Explanation:
To know the number of moles of oxygen used, first calculate the molar mass of water.
H2O=(2*1)+(1*16)
=2+16
18g/mol of H2O
The moles of H2O is 7.6 moles
So first, find the molar mass of oxygen
O2=2*16
=32g/mol
Then, number of moles of oxygen is equal to molar mass of oxygen divide by the molar mass of water times the number of moles of water
No. moles of O2=[32g/mol]/[18g/mol] *7.6moles
Moles of O2=1.778*7.6moles
No.moles of O2=13.5 moles
Therefore, the number of moles of oxygen used was 13.5moles
