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Oksanka [162]
3 years ago
11

Batteries involve the flow of electricity from one metal to another. What happens to these metals to cause the batteries to, eve

ntually, stop working?
Chemistry
1 answer:
wel3 years ago
4 0
Batteries <span>stop working due to the chemical compound between two poles running out. Also part of metal used for negative pole solved every time battery was used.</span>
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At 25.5°C a gas has a volume of 125mL. What would be the volume be if the temperature increased to 50°C
kolezko [41]

Answer:

245 mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

7 0
3 years ago
Can I get help with this, please!!
kondor19780726 [428]

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

Explanation:

Pressure in the submarine = 108.9 kPa

Volume, V = 2.4 * 10^5 L

Pressure, P = 116k Pa

Temperature, T = 312 K

Ideal gas law: PV = nRT  or  n = PV / RT

So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K

= 1.073 *10^4 mol

when temperature is changed to 293K,

PV = nRT  or P = nRT / V

=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L

=108.9 K Pa

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

8 0
2 years ago
Which is a substance that is made up of only one type of atom?
deff fn [24]
The answer is element
5 0
3 years ago
Read 2 more answers
Why do farmers spray water over their crops before a frost?
Nezavi [6.7K]

Answer:

When water freezes and turns into ice, it releases latent heat. Then, the ice that builds up on the plant will insulate it from the colder surrounding air temperatures. Because of this, some growers choose to spray their crop with water before the freeze occurs.

Explanation:

5 0
3 years ago
What is the temperature difference if 300 cal are used to warm water
PIT_PIT [208]

Answer:

<u>The temperature difference is</u> \frac{300}{m} Kkg^{-1}

Explanation:

The formula that is to used is :

ΔQ=msΔT

<em>where ,</em>

  • <em>ΔQ is the heat supplied in calories = 300cal</em>
  • <em>m is the mass of water taken = m (assumed)</em>
  • <em>ΔT is the change in temperature</em>
  • <em>s is the specific heat of water = 1calK^{-1}Kg^{-1}</em>

ΔT :

\frac{300}{m*1} \\=\frac{300}{m} Kkg^{-1}

6 0
3 years ago
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