Answer:
590 g CaCl₂
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
5.3 mol CaCl₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of CaCl₂ - 40.08 + 2(35.45) = 110.98 g/mol
<u>Step 3: Convert</u>
<u />
= 588.194 g CaCl₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules and round.</em>
588.194 g CaCl₂ ≈ 590 g CaCl₂
<span>Answer:
2Al(s) + 3Cl2(g) ---> 2AlCl3(s)
Always work in moles
moles = mass of substane / molar mass
2Al (s) + 3Cl (g) --> 2AlCl3 (s)
moles Al = 23.0 g / 26.98 g/mol
= 0.852 moles Al</span>
The volume of one mole of gas is 22.4 liters at 0 °C and 760 Torr. The volume of the given condition can be calculated using:
(PV)/T = constant
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4)/273 = (792 * V₂) / 321.7
V₂ = 25.3 Liters
The Mr of He = 4
The Mr of Ne = 20
Density of He = 4/25.3 = 0.1581 g/L
Density of Ne = 20/25.3 = 0.7905 g/L
Let x be the fraction of Ne
1 - x is the fraction of He
avg density = sum[(mole fraction * Mr)/V]
0.6048 = (20x + 4(1 - x))/25.3
x = 0.70
Answer:
The answer is growth rate
Explanation:
it will help you
The d subshell has 5 orbitals (with each being able to hold two electrons. As such, the d orbital can hold a maximum of 10 electrons.
<span>The "d" subshell can hold a maximum of _TEN_ electrons.</span>
