Answer:
- <u>No, you cannot dissolve 4.6 moles of copper sulfate, CuSO₄, in 1750mL of water.</u>
Explanation:
This question is part of a Post-Lab exercise sheet.
Such sheet include the saturation concentrations for several salts.
The saturation concentration of Copper Sulfate, CuSO₄, indicated in the table is 1.380M.
That means that 1.380 moles of copper sulfate is the maximum amount that can be dissolved in one liter of solution.
Find the molar concentration for 4.6 moles of copper sulfate in 1,750 mL of water.
You need to assume that the volume of water (1750mL) is the volume of the solution. This is, that the 4.6 moles of copper sulfate have a negligible volume.
<u>1. Volume in liters:</u>
- V = 1,750 mL × 1 liter / 1,000 mL = 1.75 liter
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<u>2. Molar concentration, molarity, M:</u>
- M = number of moles of solute / volume of solution in liters
- M = 4.6 moles / 1.75 liter = 2.6 M
Since the solution is saturated at 1.380M, you cannot reach the 2.6M concentration, meaning that you cannot dissolve 4.6 moles of copper sulfate, CuSO₄ in 1750mL of water.
1 golf ball has more mass than a tennis ball.
htdtshhthtdhthshAnswer:
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Explanation:
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
