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Arte-miy333 [17]
3 years ago
9

Calculate the following quantity: volume of 1.403 M copper(II) nitrate that must be diluted with water to prepare 575.2 mL of a

0.8012 M solution.
Chemistry
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

328 ml

Explanation:

We have given final volume =575.2 ml=0.575 L

Final concentration = 0.8012 M

We know that moles of copper(II) nitrate = final volume ×final concentration =0.8012×.0575=0.4606 moles

We have to find initial volume

So initial volume =\frac{moles\ of \ copper(II)\ nitrate}{initial\ concentration}=\frac{0.4606}{1.403}=0.328L=328ml

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Hello! Your question seems incomplete, still I will try to answer it in a general way. As shown in in equation 22 KJ heat is absorbed during the reaction so it is endothermic reaction.

Effect of Temperature:
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A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

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Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

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Answer:

Oxygen-16 (16o) is a stable isotope of oxygen, having 8 neutrons and 8 protons in its nucleus.

Explanation:

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