Answer:
190.4g
Explanation:
1.6mol of KBr (119.002g KBr/1 mol) = 190.4g
since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.
Answer:
0.85 mole of PBr3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
3Br2 + 2P —> 2PBr3
From the balanced equation above,
3 moles of Br2 reacted to produce 2 moles of PBr3.
Therefore, 1.27 moles of Br2 will react to produce = (1.27 x 2)/ 3 = 0.85 mole of PBr3.
Therefore, 0.85 mole of PBr3 is produced by the reaction.
<span>Ethylpropylamine has a chemical formula of C2H13N. It has a molecular weight of 83.166 g/mol. It's considered highly flammable and dangerous if swallowed. It should not come in contact with skin or in eyes and it should not be inhaled.</span>
Sodium lends 1 electron.
Phosphorus borrows 3 electrons.
Potassium lends one electron.
Oxygen borrows 2 electrons.
Iodine borrows one electron.
Cesium lends 1 electron.
Bromine borrows 1 electron.
Sulfur borrows 2 electrons.
And magnesium lends 2 electrons.
Answer:
Initial volume of the container (V1) = 1.27 L (Approx)
Explanation:
Given:
Number of mol (n1) = 5.67 x 10⁻²
Number of mol (n2) = (5.67 +2.95) x 10⁻² = 8.62 x 10⁻²
New volume (V2) = 1.93 L
Find:
Initial volume of the container (V1)
Computation:
Using Avogadro's law
V1 / n1 = V2 / n2
V1 / 5.67 x 10⁻² = 1.93 / 8.62 x 10⁻²
V1 = 10.9431 / 8.62
Initial volume of the container (V1) = 1.2695
Initial volume of the container (V1) = 1.27 L (Approx)