Answer:
Aluminum + Nickel(Ii) Nitrate = Nickel + Aluminum Nitrate
Explanation:
Explanation:
a. LiCl is an ionic molecule whereas water is a polar molecule with net dipole moment in it. There LiCl in water would have an ion-dipole force of interaction.
b. Both NF3 and CH3CN have dipole moment in them, since both are polar molecule. Hence, there would be dipole-dipole interaction.
c. Here both CCl4 and benzene are non polar molecules therefore, they have London dispersion force of interaction.
d. In methylamine and water both have hydrogen bonding in them. The nitrogen of CH3NH2 forms hydrogen bond with water.
Answer:
1171.12 mL
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (milliliters)
V2 = final volume (milliliters)
T1 = initial temperature (Kelvin)
T2 = final temperature (Kelvin)
According to the information provided in this question:
P1 = 300 mmHg
P2 = 140 mmHg
V1 = 400 mL
V2 = ?
T1 = 0°C = 273K
T2 = 100°C = 100 + 273 = 373K
Using P1V1/T1 = P2V2/T2
300 × 400/273 = 140 × V2/373
120000/273 = 140V2/373
120000 × 373 = 273 × 140V2
44760000 = 38220V2
V2 = 44760000 ÷ 38220
V2 = 1171.115
The new volume is 1171.12 mL
Answer:
0.0451 mol FeCl₂
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
5.72 g FeCl₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Fe - 55.85 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of FeCl₂ - 55.85 + 2(35.45) = 126.75 g/mol
<u>Step 3: Convert</u>
<u />
= 0.045128 mol FeCl₂
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.045128 mol FeCl₂ ≈ 0.0451 mol FeCl₂
Answer:
Moles of NaNO2 = 0.158
Moles of HNO2 final = 0.098
Explanation:
Given
Moles of HCl = 12
Moles of HNO2 = 0.11
Moles of NaNO2 = 0.170
HCl +NaNO2 --> HNO2 + NaCl
1 mole of HCl react with one mole of NaNO2 to produce 1 mole of NaCl and 1 mole of HNO2
Moles of NaNO2 = 0.17 - 0.012 = 0.158
Moles of HNO2 final = 0.11 - 0.012 = 0.098
