Sr is the limiting reactant.
Given the reaction equation;
2Sr + O2 (g) → 2SrO
2 moles of Sr reacts with 1 mole of O2
2 moles Sr will react with x mole of O2
x = 2 ×1/2
x = 1 mole of O2
Since we have more O2 than required, it is the reactant in excess, hence Sr is the limiting reactant.
Learn more: brainly.com/question/14225536
Answer: Statements (A), and (C) are correct.
Explanation:
The statements that are true are as follows.
- Particles in a liquid need to move more slowly in order to freeze.
When a liquid freezes the molecules get attracted towards each other. This attraction of particles occurs slowly. Hence, this statement is true.
- Attractive forces between the particles in a liquid are broken when a liquid boils.
When temperature is raised, the molecules in a liquid gains kinetic energy and start to move quickly in random directions. As a result, liquid state changes to gaseous state. Hence, this statement is true.
If the attractive force between gas molecules have to be increased, they should be moving slower instead because moving faster does not help attracting molecules together.
Hence, the statement particles in gas move fast enough to make more attractive forces when the gas condenses is not true.
When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
Answer:
1.14 M
Explanation:
Step 1: Calculate the moles corresponding to 317 g of calcium chloride (solute)
The molar mass of calcium chloride is 110.98 g/mol.
317 g CaCl₂ × 1 mol CaCl₂/110.98 g CaCl₂ = 2.86 mol CaCl₂
Step 2: Calculate the molarity of the solution
Molarity is equal to the moles of solute divided by the liters of solution.
M = moles of solute / liters of solution
M = 2.86 mol / 2.50 L = 1.14 mol/L = 1.14 M
