The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
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The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
