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alina1380 [7]
3 years ago
8

How much heat is needed to boil 45 grams of water at 100°C?​

Chemistry
1 answer:
wolverine [178]3 years ago
6 0

Answer:

Depends how much water and the temperature of the water. To heat 1 mL of water by 1 degree C 1 cal of energy (4.184 Joules) is required. Assuming that the water is at 25 degrees C, to boil one litre (liter) of water you would require 75,000 cal or 313.8 kJ.

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Explain how acid–base indicators can be used to determine the approximate pH of a substance.
mezya [45]
PH stands for potential hydrogen.
pH can be accurately tested using acid-based indicators since it is a part of the pH of something itself. (acid and bases) The indicators themselves work when the acidic properties of the indicator begins to dissolve and form ions which gives the color indicating the pH.


5 0
3 years ago
4.00 moles of CU(CN)2<br><br>Find the number of grams ​
UNO [17]

Hey there!

Cu(CN)₂

Find the molar mass.

Cu: 1 x 63.546 = 63.546

C: 2 x 12.01 = 24.02

N: 2 x 14.07 = 28.14

-----------------------------------

                      115.706 grams

The mass of one mole of Cu(CN)₂ is 115.706 grams.

We have 4 moles.

115.706 x 4 = 463

4.00 moles of Cu(CN)₂ has a mass of 463 grams.

Hope this helps!

6 0
3 years ago
If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
3 years ago
The growth rate is the death rate minus the birth rate these change because birth rates and
zysi [14]

Answer:

Hey there!

False. The growth rate is actually the birth rate minus the death rates.

Let me know if this helps :)

4 0
3 years ago
I need filling the blank spots ​
AnnZ [28]

Answer:

have you tried c

Explanation:

the chicken and I don't know if you can make it

4 0
3 years ago
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