Answer:
The answer is: <u>Al2O3</u>
Explanation:
The data they give us is:
To find the empirical formula without knowing the grams of the compound, we find it per mole:
- 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
- 0.485 g O * 1 mol O / 16 g O = 0.03 mol O
Then we must divide the results obtained by the lowest result, which in this case is 0.02:
- 0.02 mol Al / 0.02 = 1 Al
- 0.03 mol O / 0.02 = 1.5 O
Since both numbers have to give an integer, multiply by 2 until both remain integers:
Now the answer is given correctly:
Answer:
Yes it would
Explanation: Well it kinda depend on the voltage and how the battery has been in use or based on the condition
Answer: After 4710 seconds, 1/8 of the compound will be left
Explanation:
Using the formulae
Nt/No = (1/2)^t/t1/2
Where
N= amount of the compound present at time t
No= amount of compound present at time t=0
t= time taken for N molecules of the compound to remain = 4710 seconds
t1/2 = half-life of compound = 1570 seconds
Plugging in the values, we have
Nt/No = (1/2)^(4710s/1570s)
Nt/No = (1/2)^3
Nt/No= 1/8
Therefore after 4710 seconds, 1/8 molecules of the compound will be left
Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:
When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:
1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid
