The molar mass of the gas is 77.20 gm/mole.
Explanation:
The data given is:
P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams
Using the ideal Gas Law will give the number of moles of the gas. The formula is
PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole
Also number of moles is not given so applying the formula
n= mass ÷ molar mass of one mole of the gas.
n = m ÷ x ( x molar mass) ( m mass given)
Now putting the values in Ideal Gas Law equation
PV = m ÷ x RT
3.29 × 4.60 = 37.96/x × 0.08206 × 375
15.134 = 1168.1241 ÷ x
15.134x = 1168.1241
x = 1168.1241 ÷ 15.13
x = 77.20 gm/mol
If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
v = 37.9 ml
Explanation:
Given data:
Mass of compound = 1.56 kg
Density = 41.2 g/ml
Volume of compound = ?
Solution:
First of all we will convert the mass into g.
1.56 ×1000 = 1560 g
Formula:
D=m/v
D= density
m=mass
V=volume
v = m/d
v = 1560 g / 41.2 g/ml
v = 37.9 ml
Answer:
1.67mol/L
Explanation:
Data obtained from the question include:
Mole of solute (K2CO3) = 5.51 moles
Volume of solution = 3.30 L
Molarity =?
Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:
Molarity = mole of solute /Volume of solution
Molarity = 5.51 mol/3.30 L
Molarity = 1.67mol/L
Therefore, the molarity of K2CO3 is 1.67mol/L
Here is a picture of which shows you how many valence electrons are in the Lewis structure of xeo4