This reaction involves two soluble ionic compounds and so is a potential double-displacement reaction. Follow this procedure to write balanced molecular equations for double displacement reactions, when you know only the formulas of the reactants
Answer:
678 mL.
Explanation:
The balanced reaction between calcium chloride and silver nitrate is shown in figure.
As shown in the equation , each mole of aqueous calcium chloride will rect with two moles of silver nitrate.
The moles of calcium chloride taken = molarity X volume (L)
= 0.37X0.33 = 0.122 mol
the moles of silver nitrate required = 2 X 0.122 = 0.244 mol
the volume of silver chloride required =
M = 0.36 M
mol = 0.244
Putting values
volume = 0.678 L = 678 mL.
Answer:
Carbon Monoxide / Carbon Dioxide / Sulfur and Nitrogen Dioxide
Explanation:
<h3>
Answer:</h3><h3>a) 9.033 × 10²³ particles</h3><h3>b) 4.068 × 10²⁴ particles</h3><h3>c) 1.51 × 10²³ particles</h3>
Explanation:
For us to answer these questions, we have to know two formulas:
- Number of particles = moles × Avogadro's Number
- Moles = Mass ÷ Molar Mass
Therefore:
a) particles of Na = 1.50 mol × (6.022 × 10²³) particles/mol
= 9.033 × 10²³ particles
b) particles of Pb = 6.755 mol × (6.022 × 10²³) particles/mol
= 4.068 × 10²⁴ particles
c) particles of Si
= (7.02 g ÷ 28.085 g/mol) × (6.022 × 10²³) particles/mol
= 1.51 × 10²³ particles
Answer:
V(NH₃(g)) at STP = 0.8928 Liters ≅ 0.9 Liter (1 sig. fig)
Explanation:
Given 340g NH₃(g) => ?Volume at STP
Convert grams to moles by dividing by formula wt of ammonia
= 340g/17g/mole = 20 moles
1 mole of any gas at STP conditions (0°C, 1atm) occupies 22.41 Liters
∴ 20 moles NH₃(g) => 20moles/22.4Liters/mole
= 0.8928 Liters ≅ 0.9 Liter (1 sig. fig)