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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Answer:
The correct answer is - 4.
Explanation:
As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -
The sum of the product mass number of products = mass of reactant
237Np93 →233 Pa91 +AZX is the equation,
Solution:
Mass of reactants = 237
Mass of products are - Pa =233 and A = ?
233 + A = 237
A = 237 - 233
A = 4
So the equation will be:
237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)
Answer:
B. Charges ( a slight positive charge on one end, and a slight negative charge on the other).
Answer:
When you have the feeling of danger and you either don't want to face it or you want to face it.
Explanation:
Answer:- B. 4.65 g.
Solution:- The given balanced equation is:
It asks to calculate the mass of silver sulfide formed by when 0.0150 liters of 2.50 M of silver nitrate are used.
Moles of silver nitrate are calculated on multiplying it's liters by its molarity and then on multiplying by mol ratio, the moles of silver sulfide are calculated. These moles are multiplied by the molar mass to convert to the grams.
Molar mass of 
= 2(107.87)+32.06 = 247.8 g per mol
The dimensional set up for the complete problem is:
= 
So, the correct choice is B. 4.65 g.
