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ale4655 [162]
3 years ago
6

An experiment calls for you to use 100 mL of 0.25 M HNO3 solution. All you have available is a bottle of 3.4 M HNO3. How many mi

lliliters of the 3.4 M HNO3 solution do you need to prepare the desired solution?
Chemistry
1 answer:
uysha [10]3 years ago
3 0

Answer: 7.35mL

Explanation:

C1 = 3.4M

V1 =?

C2 = 0.25M

V2 = 100mL

C1V1 = C2V2

3.4 x V1 = 0.25 x 100

V1 = (0.25 x 100) /3.4

V1 = 7.35mL

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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!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

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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