Answer:
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Explanation:
<h2>#CarryOnLearning</h2>
Answer:
Option D is good to go!
Explanation: as per the reactivity series more reactive substances will react with the counterpart substance.The most reactive substance here is calcium while the least reactive is aluminium, the magnesium comes in between.As per their reactivity, these substances will react with oxygen.
Explanation:
Answer:
525.1 g of BaSO₄ are produced.
Explanation:
The reaction of precipitation is:
Na₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) ↓ + 2NaCl (aq)
Ratio is 1:1. So 1 mol of sodium sulfate can make precipitate 1 mol of barium sulfate.
The excersise determines that the excess is the BaCl₂.
After the reaction goes complete and, at 100 % yield reaction, 2.25 moles of BaSO₄ are produced.
We convert the moles to mass: 2.25 mol . 233.38 g/mol = 525.1 g
The precipitation's equilibrium is:
SO₄⁻² (aq) + Ba²⁺ (aq) ⇄ BaSO₄ (s) ↓ Kps
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
