Answer:
mass of CO = 210.42 g
mass in three significant figures = 210. g
Explanation:
Given data:
mass of Fe2O3 = 0.400 Kg
mass of CO= ?
Solution:
chemical equation:
Fe2O3 + 3CO → 2Fe + 3CO2
Now we will calculate the molar mass of Fe2O3 and CO.
Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol
Molar mass of CO = 12+ 16 = 28 g/mol
now we will convert the kg of Fe2O3 in g.
mass of Fe2O3 = 0.400 kg × 1000 = 400 g
number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol
mass of CO = moles of Fe2O3 × 3( molar mass of CO)
mass of CO = 2.505 mol × 84 g/mol
mass of CO = 210.42 g
mass in three significant figures = 210. g
Answer:
16.89g of PbBr2
Explanation:
First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:
Molarity of Pb(NO3)2 = 0.595M
Volume = 77mL = 77/1000 = 0.077L
Mole =?
Molarity = mole/Volume
Mole = Molarity x Volume
Mole of Pb(NO3)2 = 0.595x0.077
Mole of Pb(NO3)2 = 0.046mol
Convert 0.046mol of Pb(NO3)2 to grams as shown below:
Molar Mass of Pb(NO3)2 =
207 + 2[ 14 + (16x3)]
= 207 + 2[14 + 48]
= 207 + 2[62] = 207 +124 = 331g/mol
Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g
Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol
Equation for the reaction is given below:
Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2
From the equation above,
331g of Pb(NO3)2 precipitated 367g of PbBr2
Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2
Answer: he gave him a map of somewhere
Explanation:
Answer:
0.125 moles
Explanation:
2.8 litres is equivalent to 2.8dm³
At STP,
1 mole = 22.4 dm³
x mole = 2.8 dm³
Cross multiply
22.4x = 2.8
Divide both sides by 22.4
x = 2.8/22.4
x = 0.125
