Answer:
The final pressure of the gas is:- 21.3 kPa
Explanation:
Using Boyle's law
Given ,
V₁ = 10.0 L
V₂ = 45.0 L
P₁ = 96.0 kPa
P₂ = ?
Using above equation as:
The final pressure of the gas is:- 21.3 kPa
78.4 L volume of container is required to hold 3.2 moles of gas.
Explanation:
- STP is defined as the standard temperature and pressure of a gas in room temperature conditions. At STP, one mole of the gas which has Avogadro's number of molecules in it will occupy a volume of 22.4 L.
- So, one mole of a substance or gas will occupy a volume of 22.4 L then the volume of the container needed for 3.2 moles of gas is calculated by multiplying 22.4 L, standard volume with the moles of the gas 3.2 moles.
- Hence, the answer would be 78.4 L.
Answer:
the correct answers is 100 22.7 and 24.6
Explanation:
did it on edgunity
Additional information
Relative atomic mass(Ar) : A=7, G=16
The empirical formula : A₂G
<h3>Further explanation</h3>
Given
3.5g of element A
4.0g of element G
Required
the empirical formula for this compound
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
The empirical formula also shows the simplest mole ratio of the constituent elements of the compound
mol of element A :
mol of element G :
mol ratio A : G = 0.5 : 0.25 = 2 : 1
If 4 moles of P is used by 5 mole of O2
then....0.489 moles will be used by 5/4 × .489 = .611 moles of O2
so .611 moles
so if 4 moles of P is burnt , 1 mole of P4O10 is produced ....so for .489 moles...... .489/4=.122 moles !
so mass will be .122× 283.89 = 34.7 grams
so first ans is .611 moles and second is 34.7 grams !
if you have any problem regarding this , just comment !!!
