The balanced chemical equation is written as:
<span>CsF(s) + XeF6(s) ------> CsXeF7(s)
We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:
11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
1) Hydrocarbon: CH3 - CH2 - CH2 - CH2 - CH3
2) Only single bonds => alkane => sufix ane
3) no substitutions
4) 5 carbons = > prefix penta.
Therefore, the name is pentane.
Answer:
[CaCl₂·2H₂O] = 1.43 m
Explanation:
Molality is mol of solute / kg of solvent.
Mass of solvent = 40 g
Let's convert g to kg → 40 g / 1000 = 0.04 kg
Let's determine the moles of solute (mass / molar mass)
8.43 g / 146.98 g/mol = 0.057 mol
Molality = 0.057 mol / 0.04 kg → 1.43
It is by looking on the dates of the paper.
Hoped this helped.
~Bob Ross®
Answer:
400 mL
Explanation:
Given data:
Mass of barium = 2.17 g
Pressure = 748 mmHg (748/760 = 0.98 atm)
Temperature = 21 °C ( 273+ 21 = 294k)
Milliliters of H₂ evolved = ?
Solution:
chemical equation:
Ba + 2H₂O → Ba(OH)₂ + H₂
Number of moles of barium:
Number of moles = mass/ molar mass
Number of moles = 2.17 g / 137.327 g/mol
Number of moles = 0.016 mol
Now we will compare the moles of barium with H₂.
Ba : H₂
1 : 1
0.016 : 0.016
Milliliters of H₂:
PV = nRT
V = nRT/P
V = 0.016 mol × 0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm
V = 0.39 atm. L/0.98 atm
V = 0.4 L
L to mL
0.4 × 1000 = 400 mL
