Answer:
648.5 mL
Explanation:
Here we will assume that the pressure of the gas is constant, since it is not given or specified.
Therefore, we can use Charle's law, which states that:
"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"
Mathematically:
where
V is the volume of the gas
T is its absolute temperature
The equation can be rewritten as
where in this problem we have:
is the initial volume of the gas
is the initial temperature
is the final temperature
Solving for V2, we find the final volume of the gas:
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion
1) Content of Ca (2+) ions
Calcium chloride = CaCl2
Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)
=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)
Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution
M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2
=> 0.0825 mol Ca(2+)
2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)
formula of phospahte ion: PO4 (3-)
molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2
Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)
=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)
3) Content of Mg(2+) ions
Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)
Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)
number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution
n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2
ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)
4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)
2PO4(3-) + 3Mg(2+) = Mg3(PO4)2
=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)
=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)
5) Total number of moles of PO4(3-)
0.055 mol + 0.16 mol = 0.215 mol
6) Sodium phosphate
Sodium phosphate = Na3(PO4)
Na3PO4 ---> 3Na(+) + PO4(3-)
=> 1 mol Na3PO4 : 1 mol PO4(3-)
=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4
mass in grams = number of moles * molar mass
molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol
=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g
Answer: 35.26 g of sodium phosphate
</span>
To Find :
The volume of 12.1 moles hydrogen at STP.
Solution :
We know at STP, 1 mole of gas any gas occupy a volume of 22.4 L.
Let, volume of 12.1 moles of hydrogen is x.
So, x = 22.4 × 12.1 L
x = 271.04 L
Therefore, the volume of hydrogen gas at STP is 271.04 L.
Answer:
0.17%
Explanation:
With the equation:
2Cr2O7 2- + C2H5OH + H2O --> 4Cr3+ + 2CO2 + 11H2O
We can assume that every mole of ethanol needs 2 moles of Dichromate to react.
So if in 1L we have 0.05961 moles of dichromate we can discover how many moles we have in 35.46mL
1000 mL - 0.05962 moles
35.46 mL - x
x = 
x = 2,11* 10^-3 moles
As we said earlier, 1 mole of ethanol needs 2 mole of dichromate, so in the solution we have 1,055*10^-3 moles of ethanol. We can discover the mass of ethanol present in the solution.
1 mole - 46g
1.055*10^-3 - y
y = 46 * 1.055*10^-3
y = 0.048 g
To discover the percent of alchol we can use a simple relation
28 g - 100%
0.048 - z
z = 
z = 0.17%
