Answer:
52.0 g of LiCl represents 1.23 moles of compound
Explanation:
In order to determine the moles of LiCl, we need to know the molar mass of our compound
Molar mass LiCl = Molar mass Li + Molar mass Cl
Molar mass LiCl = 6.94 g/mol + 35.45 g/mol =42.39 g/mol
Now, we can do the conversion:
52 g . 1 mol / 42.39 g/mol = 1.23 moles
Mass of water = 20.88 - 16.51
Mass of water = 4.37 g
Percentage water in hydrate = 4.37 / 20.88 x 100
Percentage water inn hydrate = 20.9%
Answer:
18
Explanation:
The atomic number of chlorine is 17 therefore the atom has 17 electrons, the ion however has 1 extra electron. The charge becomes negative because electrons have a negative charge. 17 + 1 = 18, therefore a chlorine ion with a charge of -1 has 18 electrons.
<u>Answer:</u>
<em>The percentage error in the measurement is 2.4%
</em>
<u>Explanation:</u>
The measurement made by the student is 
.
But the actual value of density is 
The formula to calculate percentage error is
percentage error = 
Hence percentage error in this case is given by
<em>Percentage error = 
</em>
<em>= 
</em>
A physical quantities value will be determines by various quantities. If errors are made in measuring any of these quantities there will be error in the measurement of the final physical quantity as well.
Here density of a substance is a derived quantity determined by measuring mass and volume of that substance.<em> If errors are made in mass and volume measurements the density value will also be erroneous.
</em>
Answer:
The answer to your question is:
Vol of NO2 = 11.19 L
Vol of O2 = 2.8 L
Explanation:
Data
N2O5 = 56 g
STP T = 0°C = 273°K
P = 1 atm
MW N2O5 = 216 g
Gases law = PV = nRT
Process
216 g of N2O5 ---------------- 1 mol
54 g ----------------- x
x = (54 x 1) / 216
x = 0.25 mol of N2O5
2 mol of N2O5 ----------------- 4 mol of NO2
0.25 mol ------------------ x
x = (0.25 x 4) / 2 = 0.5 mol of NO2
V = nRT/P
V = (0.5)(0.082)(273) / 1 = 11.19 L
2 mol of N2O5 ----------------- 1 O2
0.25 N2O5 ---------------------- x
x = (0.25 x 1) / 2 = 0.125 mol
Vol = (0.125)((0.082)(273) / 1 = 2.8 L
