Answer:
0.0071g
Explanation:
From the question, we know that the molarity of the BaCl2 is 0.00237M. This means there are 0.00237 moles in 1dm^3 or 1000cm^3 of solution.
We also know that 35ml of the BaCl2 reacted. Here, we need to calculate the number of moles in 35.7ml of BaCl2.
This is calculated as follows;
0.00237moles are in 1000cm^3
Thus x moles will be present in 35ml ( we should note that cm^3 is same as ml)
X = (0.00237 × 35) ÷ 1000 = 0.00008295 moles.
From the reaction equation, we can see that 2 moles of BaCl2 yielded 1 mole of Ba(OH)2.
This means 0.00008295mole of BaCl2 will yield 0.00008295 ÷ 2 = 0.000041475 moles of Ba(OH)2.
To calculate the mass of Ba(OH)2 formed, we simple multiply the number of moles yielded by the molar mass of Ba(OH)2.
Molar mass of Ba(OH)2 = 137 + 2(17)
= 171g/mol
Mass = 171 × 0.000041475 = 0.007092225g
Answer:
726 torr
Explanation:
Generally, atmospheric pressure can be measured using a manometer which is in form of a U-shaped tube. In addition, 1 mm Hg is equivalent to 1 torr. Therefore, 752 torr is equivalent to 752 mm Hg. Therefore, the total pressure will be equivalent to the atmospheric pressure (mm Hg) + the mercury height.
In this case, the mercury height = -26 mm
Thus:
The helium pressure = 752 - 26 = 726 mm Hg
This is also equivalent to 726 torr
If the reaction is represented by:
PCl₃ + Cl₂ <-> PCl₅ (exothermic)
the mole fraction of chlorine in the equilibrium mixture will change according to the following:
Decrease the volume: decrease
Increase the temperature: increase
Increase the volume: increase
Decrease the temperature: decrease
On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol
18g C*1mol C/12 g C = 1.5 mol C
C + O2 → CO2
from reaction 1 mol 1 mol 1 mol
from problem 1.5 mol 1.5 mol 1.5 mol
1.5 mol O2*32 g O2/1 mol O2 = 48 g O2
In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount of O2 that was added for the reaction (probably people who wrote this problem wanted this)
we need (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is </span><span>2.) 54 g.</span>
