Events that happen over and over create/become a pattern
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen 
if it occupies 31.6 mL at 
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
Temperature = 
As molar mass of 
is 32 g/mol. Hence, the number of moles of are calculated as follows.
Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen if it occupies 31.6 mL at .
A. is decomposition so HCL = H2 + Cl2
not balanced cause hcl needs 2
2HCL = H2 + Cl2
balanced
b. Br2 + Al-i = AlBr3 + I2 single rep.
not balanced since br need 3 so watch carefully cause many changes needed
3Br2 + Al-i = AlBr3 + I2 not right is unbalanced so make it 2
3Br2 + Al-i = 2AlBr3 + I2 now left Al is unbal. so make 2 there
3Br2 + 2Ali = 2AlBr3 + I2
Balanced
C. Na + S = Na2S synthesis reaction is not bal. left Na needs 2
2Na + S = Na2S balanced.
To solve this problem, we assume ideal gas so that we can
use the formula:
PV = nRT
since the volume of the flask is constant and R is
universal gas constant, so we can say:
n1 T1 / P1 = n2 T2 / P2
1.9 mol * (21 + 273 K) / 697 mm Hg = n2 * (26 + 273 K) /
841 mm Hg
<span>n2 = 2.25 moles</span>
Answer:
Explanation:
a )
1.25 g MgO contains .754 g of Mg .Rest will be O
so oxide = 1.25 - .754 = 0.496 g
ratio of magnesium to oxide = .754/.496 = 1.52
b) 1.25 g of MgO contains .754 g of Mg
534 g of MgO contains .754 x 534 / 1.25 g = 322.11 g
