How many joules of heat are absorbed when 73 g water are heated from 30*C to 43*C? *
2 answers:
Answer:
Explanation:
We are given the mass and change in temperature, so we must use this formula for heat energy:
The mass is 73 grams. Water's specific heat is 4.18 J/g × °C. Let's calculate the change in temperature
- ΔT= final temperature - initial temperature
- ΔT= 43 °C - 30°C
- ΔT= 13 °C
Now we know all the variables and can substitute them into the formula.
Multiply the first numbers together. The grams will cancel.
Multiply again. This time, the degrees Celsius cancel.
<u>39,668.2 Joules</u> of heat energy are absorbed.
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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!