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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If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.
It is due to the law of conservation of mass.
Mass can neither be created nor be destroyed.
A water molecule, because of its shape, is a polar molecule. That is, it has one side that is positively charged and one side that is negatively charged. The molecule is made up of two hydrogen atoms and one oxygen atom. The bonds between the atoms are called covalent bonds, because the atoms share electrons.
Answer:
Inhibition and tolerance.
Explanation:
Competitive inhibitors are characterized by being chemicals that are similar to the normal substrate. While non-competitive inhibitors are chemicals that bind to the enzyme and alter its shape. This renders the active site of the enzyme ineffective against the substrate. Tolerance makes a society that is fair and equitable.
Answer:
465mL
Explanation:
Volume of a solution, V =Mass of substance, m/(Molarity of the solution of the substance, M × molar mass of the substance, M.m)
Given in the question,
M=.132M
M.m=23+35.5 = 58.5g/mol
m=3.59g
V= 3.59/(.132×58.5)
V = 0.465L
Volume in mL = volume in L × 1000
= 0.465 × 1000 = 465mL
Therefore, 465mL of a .132M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt
