Answer:
ΔT = 1,41°C
Explanation:
For the reaction:
2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2 H₂O(l); ΔH = –118 kJ.
If moles of reaction are:
HCl:
0,0500L×0,20M = 0,01moles
Ba(OH)₂:
0,0500L×0,10M = 0,005moles
The heat produced is:
0,005mol×-118 kJ/mol = -0,59 kJ = -590 J
The temperature increasing is:
Q = -C×m×ΔT
Where Q is heat (-590 J), m is mass of solution (100mL = 100g -density of 1,00g/mL) and ΔT is change in temperature.
Replacing:
-590J = -4,184J/g°C×100g×ΔT
<em>ΔT = 1,41°C</em>
<em></em>
I hope it helps!
Answer:
The final molarity of acetate anion in the solution is 0.0046 moles
Explanation:
The balanced equation is
Cu(C₂H₃O₂)₂ + Na₂CrO₄ = CuCrO₄ + 2Na(C₂H₃O₂)
Therefore one mole of Cu(C₂H₃O₂)₂ react with one mole of Na₂CrO₄ to form one mole of CuCrO₄ and two moles of Na(C₂H₃O₂)
Mass of copper (II) acetate present = 0.708 g
Volume of aqueous sodium present = 50 mL
Molarity of sodium chromate = 46.0 mM
Therefore
Number of moles of sodium chromate present = (50 mL/1000)×46/1000 = 0.0023 M
Number of moles of copper (II) acetate present = 181.63 g/mol
number of moles of copper (II) acetate present = (0.708 g/181.63 g/mol) =0.0039 moles
Therefore 0.0039 moles of Cu(C₂H₃O₂)₂ × (2 moles of Na(C₂H₃O₂))/1 Cu(C₂H₃O₂)₂) = 0.00779 moles of Na(C₂H₃O₂)
also 0.0023 moles of Na₂CrO₄ × (2 moles of Na(C₂H₃O₂))/1 Na₂CrO₄) = 0.0046 moles of Na(C₂H₃O₂)
Therefore the Na₂CrO₄ is the limiting reactant and 0.0046 moles of Na(C₂H₃O₂) or acetate anion is formed
Answer:
1.070MKCl
Explanation:
So we know that the original formula is M= n/L (n being moles of solute, L being liters of solvent)
Since we do not have liters in this problem, we would need to convert milliliters to liters
<u>213 mL= 0.213 L</u>
We then see that we do not have moles, but we do have a mass, being <u>17.0 g.</u> we would need to convert these grams to moles, giving us <u>0.228 mol.</u>
Then, you would plug in <u>0.228 for your n</u>, and now you are ready to solve your original formula, plugging everything in.
M=n/L
M=0.228 mol/0.213L
M= 1.070MKCl
I know this was long, but I hope this helps (:
The elements of group 1 makes ionic bond with the elements of group 7 due to high difference of electronegativity values.
<h3>Type of bond between group 1 and 7</h3>
The elements of first group lose its one outermost electrons while on the other hand, the elements of seven group needs one electron so they gain that one electron so they make an ionic bond with each other.
So we can conclude that the elements of group 1 makes ionic bond with the elements of group 7 because of the high difference of electronegativity values.
Learn more about electronegativity here: brainly.com/question/2415812
Answer:i
i believe its contraction but i might be wrong
Explanation:
