Answer:
I think it will be (B) only because you have to do it repeatedly and analyze it and not noted, in order for the results are accepted. I May be wrong. Thanks for letting me help you.
830 mL
The volume of an 2.3 m solution with 212 grams of calcium chloride (cacl2) dissolved is 830 mL.
The solution has a concentration of 2.3 mol/L.
<h3>a) Moles of CaCl2</h3>
Molar mass of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)
= 1.910 mol CaCl2
<h3>b) Volume of solution</h3>
V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution
= 830 mL solution
<h3>How much CaCl2 is there in the solution by molarity?</h3>
- The number of moles is 0.125 x 2 = 0.25 mol since the molarity is 2.0M.
- To get the answer of 27.745 g, simply multiply this by the molar mass of calcium chloride, which is 110.98 g/mol.
To learn more about CaCl2 solution visit:
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I would say D and if that's not it it's A
Hope This Helps! Have A Nice Day!!
Answer:
Final temperature = T₂ = 328.815 K
Explanation:
Given data:
Given energy = 980 KJ = 980×1000= 980000 J
Volume = 6.2 L
Initial temperature =T₁= 291 K
Specific heat of water = 4.18 j /g .K
Final temperature = T₂ = ?
Formula:
Q = m. c . ΔT
ΔT = T₂ - T₁
we will first convert the litter into milliliter
6.2 × 1000 = 6200 mL
It is given in question that
1 mL = 1 g
6200 mL = 6200 g
Now we will put the values in formula,
Q = m. c . (T₂ - T₁)
980000 j = 6200 g . 4.18 j /g .K . (T₂ - 291 K)
980000 j = 25916 j/ k . (T₂ - 291 K)
980000 j / 25916 j/ k = T₂ - 291 K
37.8145 K + 291 K =T₂
T₂ = 328.815 K
