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Lemur [1.5K]
3 years ago
12

What does kcal mean again?

Chemistry
2 answers:
sasho [114]3 years ago
8 0

Answer:

1000 Calories/kilocalorie/large calorie/food calorie

Explanation:

kcal

K comes form the Greek kilo and is the SI prefix for 1000

cal means calorie

so kcal or kilocalorie or large calorie or food calorie

The large calorie, food calorie, or kilocalorie (Cal, calorie or kcal) is the amount of heat needed to cause a one degree Celsius temperature increase in a kilogram of water.

Neporo4naja [7]3 years ago
4 0
It means kilocalorie
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Is o-toluic acid soluble in ether, NaOH?, Is Napthalene soluble in NaOH?
natima [27]
Yes, o-toluic acid is soluble in ether as ether is slightly polar and it is soluble in NaOH because it is likely to form soluble compounds with it.

Naphthalene is insoluble in NaOH.
8 0
3 years ago
12.00 moles of naclo3 will produce how many grams of O2? (Moles to Grams)
Sergeu [11.5K]
2NaClO₃   →  2NaCl  +  3O₂

mole ratio of NaClO₃  to  O₂  is  2  :  3

∴  if moles of NaClO₃  =  12 mol

then moles of O₂  =  \frac{12 mol   *   3}{2}
      
                             =  18 mol


Mass of O₂  =  mol of O₂  ×  molar mass of O₂
 
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So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction.  If you should have used another reaction then just follow the same steps and you'll get your answer.




4 0
3 years ago
A little boy places an ice cube in his hot bowl of stew and then walks out of the room. When he returns, the ice cube is gone.
Alenkinab [10]

Answer:

b

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3 0
3 years ago
How do cells get ATP, the energy currency that does work in living things?
luda_lava [24]
How do cells get ATP through Cellular Respiration.
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3 0
3 years ago
A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached
lara [203]

Answer:

a)  cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b)  0.068 V.

Explanation:

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Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

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E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

 = 0.339V + (90.05916V/2)log(0.100) = 0.309V

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6 0
3 years ago
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