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: +178.3 kJ
Explanation:
The chemical equation follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaO%28s%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5E0f_%7BCO_2%7D%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaCO_3%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-635.1%29%29%2B%281%5Ctimes%20%28-393.5%29%29%5D-%5B%281%5Ctimes%20%28-1206.9%29%29%5D)
The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ
From Ohm's law, we get the general equation that would relate the voltage, current, and resistance,
V = I x R
where V is voltage, R is resistance, and I is current. Deriving the equation for R
R = V / I
R = (1.5 volts) / (0.7 amps) = 2.14 Ohms
Answer:
Explanation:
Atoms are the thing that make up molecules and compounds. Molecules contain two or more atoms and are held together by covalent bonds
