Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
Answer:
HELLO THERE!
I HOPE MY ANSWER WILL HELP YOU :)
Explanation:
There is this picture that helped me. I hope it helps you too.
Answer:
XZ2
Explanation:
There are different ways in which compounds can be represented. Broadly, we have three different types of formula;
- Structural formular: This shows how th atoms in te compound are connected to each other.
- Molecular formular: This shows the actual number of atoms of element present in the compound
- Empirical Formular: This is the simplest formular of a compound. It basically shows the number of atoms in simple ration to each other.
This question requires us to input the empirical formular;
X2Z4
The ratio of the elements is; 2 : 4 which can be simplified into 1 : 2
This means the empirical formular is XZ2
Answer:- 13.6 L
Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.
Standard pressure is 1 atm that is 101.325 Kpa.
Boyle's law equation is:

From given information:-
= 58.7 Kpa
= 23.5 L
= 101.325 Kpa
= ?
Let's plug in the values and solve it for final volume.

On rearranging the equation for 

= 13.6 L
So, the volume of hydrogen gas at STP for the given information is 13.6 L.