The first ionization energy of a known element is the energy
it needs to remove its highest energy or outermost electron. It is done to make
a neutral atom be a positively charged ion. The first ionization energy of neon
as a chemical equation is this:
Ne (g) -> Ne+ (g) + e-
Answer:
<u>1. Net ionic equation:</u>
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s)
<u />
<u>2. Volume of 1.0M AgNO₃</u>
Explanation:
1. Net ionic equation for the reaction of NaCl with AgNO₃.
i) Molecular equation:
It is important to show the phases:
- (aq) for ions in aqueous solution
- (s) for solid compounds or elements
- (g) for gaseous compounds or elements
- NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)
ii) Dissociation reactions:
Determine the ions formed:
- NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
- AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
- NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)
iii) Total ionic equation:
Substitute the aqueous compounds with the ions determined above:
- Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
iv) Net ionic equation
Remove the spectator ions:
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer
2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl
i) Determine the number of moles of AgNO₃
The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl
The number of moles of AgCl is determined using the molar mass:
- number of moles = mass in grams / molar mass
- molar mass of AgCl = 143.32g/mol
- number of moles = 5.84g / (143.32g/mol) = 0.040748 mol
ii) Determine the volume of AgNO₃
- molarity = number of moles of solute / volume of solution in liters
- V = 0.040748mol / (1.0M) = 0.040748 liter
- V = 0.040748liter × 1,000ml / liter = 40.748 ml
Round to two significant figures: 41ml ← answer
Explanation:
Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.
Mass of sample (CuSO4 * nH2O)
= 152.00g - 128.10g = 23.90g.
Mass of water loss during heating
= 152.00g - 147.60g = 4.40g.
Molar mass of H2O = 18g/mol
Moles of H2O in sample
= 4.40g / (18g/mol) = 0.244mol.
Mass of anhydrous sample (CuSO4)
= 23.90g - 4.40g = 19.50g
Molar mass of CuSO4 = 159.61g/mol
Moles of CuSO4 in sample
= 19.50g / (159.61g/mol) = 0.122mol.
Since mole ratio of CuSO4 to H2O
= 0.122mol : 0.244mol = 1:2, n = 2.
Hence we have CuSO4 * 2H2O.
Answer:
The molarity of this sugar solution in water is 2.18 M
Explanation:
Step 1: Data given
Mass of sugar (C12H22O11) = 186.55 grams
Molar mass of C12H22O11 = 342.3 g/mol
Volume of water = 250.0 mL = 0.250 L
Step 2: Calculate moles sugar
Moles sugar = mass sugar / molar mass sugar
Moles sugar = 186.55 grams / 342.3 g/mol
Moles sugar = 0.545 moles
Step 3: Calculate molarity of the sugar solution
Molarity = moles sugar / volume of water
Molarity = 0.545 moles / 0.250 L
Molarity = 2.18 MThe molarity of this sugar solution in water is 2.18 M
Hey there!:
Molar mass Ca(NO2)2 = 132.089 g/mol
Mass of solute = 120 g
Number of moles:
n = mass of solute / molar mass
n = 120 / 132.089
n = 0.0009084 moles of Ca(NO2)2
Volume in liters of solution :
240 mL / 1000 => 0.24 L
Therefore:
Molarity = number of moles / volume of solution
Molarity = 0.0009084 / 0.24
Molarity = 0.003785 M
Hope that helps!