How many mL of 0.506 M HClare needed to dissolve 9.85 g of BaCO3?
1 answer:
Answer:
197mL of 0,506M HCl
Explanation:
The reaction of HCl + BaCO₃ is:
BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.
The moles of BaCO₃ in 9,85 g are:
9,85 g of BaCO₃ × = <em>0,0499 moles of BaCO₃</em>
As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:
0,0499 moles of BaCO₃ × =<em> 0,0998 moles of HCl</em>
If you have a 0,506M HCl, you need to add:
0,0998 moles of HCl× = 0,197 L ≡ 197mL
I hope it helps!
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Answer:
x = 0.324 M s⁻¹
Explanation:
Equation for the reaction can be represented as:
2 NO(g) + Cl₂ (g) ⇄ 2NOCl (g)
Rate = K [NO]² [Cl₂]
Concentration =
from the question; their number of moles are constant since the species are quite alike.
As such; if Concentration varies inversely proportional to the volume;
we have: Concentration ∝
Concentration =
Similarly; the Rate can now be expressed as:
Rate = K [NO]² [Cl₂]
Rate =
Rate =
We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹
So we can have:
0.0120 =
0.0120 = -----Equation (1)
Now; the new rate of formation when the volume of the system decreased to 1.00 L can now be calculated as:
x =
x = 1 ------- Equation (2)
Dividing equation (2) with equation (1); we have:
=
=
x = 0.0120 × 27
x = 0.324 M s⁻¹
∴ the new rate of formation of NOCl = 0.324 M s⁻¹