I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture.
Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
<em>Q = +6970 kJ</em>
Answer:
True
Explanation:
learned it from class on a video
Answer:
0.42 M
Explanation:
The reaction that takes place is:
- Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)
First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:
(200 mL = 0.200L)
- 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄
Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:
- 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂
Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:
- 0.224 mol - 0.14 mol = 0.085 mol
Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:
- 0.085 mol / 0.200 L = 0.42 M
Answer:

Explanation:
We will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 315.46 76.12
Ba(OH)₂·8H₂O + 2NH₄SCN ⟶ Ba(SCN)₂ + 2NH₃ + 10H₂O
m/g: 6.5
1. Moles of Ba(OH)₂·8H₂O

2. Moles of NH₄SCN
The molar ratio is 2 mol NH₄SCN:1 mol Ba(OH)₂·8H₂O

3. Mass of NH₄SCN

Anything in group 18. These are the noble gases, known for having the stable octet in the outermost electron shell (with the exception of Helium). These elements include Neon, Xenon, and Radon, to name a few.