Answer:
621.41 grams
Explanation:
First of all, to find the mass. You have to use the molar mass of AlBr3 to convert from moles to mass.
Second of all, find the molar mass of AlBr3.
Al = 27.0 amu
Br = 79.9 amu
They are 3 bromine atoms and 1 aluminum atom.
So you do this
27.0 + 79.9(3) = 266.7
The molar mass is 266.7 g/mol
Third of all, use dimensional analysis to show your work.
2.33 moles of AlBr3 * 266.7 g/mol / 1 mol
Moles cancel out.
2.33 x 266.7 = 621.411
Your teacher though said to round digits after the decimal point instead of using sig figs.
621.411 rounds to 621.41
So the final answer is 621.41 grams(don't forger the units).
The substance formula is just AlBr3.
That's all.
Hope it helped!
Answer:
no it is a ionic compound
Answer:
276g of Br are present
Explanation:
To solve this question we need to find the moles of CaBr2 using its molar mass -CaBr2: 199.89g/mol-. As 1 mole of CaBr2 contains 2 moles of Br we can find the moles of Br and its mass:
<em>Moles CaBr2:</em>
345g * (1mol/199.89g) = 1.7259 moles CaBr2
<em>Moles Br:</em>
1.7259 moles CaBr2 * (2mol Br / 1mol CaBr2) = 3.4519 moles Br
<em>Mass Br -Molar mass: 79.904g/mol-</em>
3.4519 moles Br * (79.904g/mol) = 276g of Br are present
Answer:
0.031 parts per million
Explanation:
80 micrograms/m^3 = 80 micrograms/m^3 × 1m^3/1000L = 0.08 micrograms/L
Concentration in parts per million = concentration in micrograms/L × molar volume/MW
Concentration in micrograms/L = 0.08
Molar volume at 25°C and 101.325kPa (1 atm) is 24.45L
MW of SO2 = 64g/mole
Concentration in ppm = 0.08×24.45/64 = 0.031 ppm
