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seraphim [82]
3 years ago
8

If 5.40 kcal of heat is added to 1.00 kg of water at 100⁰C, how much steam at 100⁰C is produced? Show all calculations leading t

o an answer.
Chemistry
2 answers:
Eddi Din [679]3 years ago
8 0
Heat of vaporization of water = 540 cal/g
m=1000 g
energy supplied = 5.4 kcal=5400 cal
T1=T2=100 deg. C
so all energy is used to convert to steam
mass of steam produced = 5400 cal /(540 cal/g)
=5400/540 cal *g/cal
=10 g
lora16 [44]3 years ago
7 0

A kind of transformation known as phase change, in this when one is performing phase change, the component remains at the similar temperature all through the phase change. This is illustrated as whether one is heating or cooling the substance down. The heat subtracted or added relies upon the mass of the substance or the Latent Heat of the substance.  

The heat added or subtracted is given by the formula:  

Q = mL

Q is heat, L is latent heat of vaporization, which is 539 kcal/kg, and m is the mass of steam

m = Q/L = 5.4/539 = 0.01 Kg

m = 10 grams of steam.  


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How many grams of ammonia produced from 1000 grams of N2?
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N2 + 3H2 ———> 2NH3

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5 0
3 years ago
Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of
Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles

Now we have to calculate the moles of PbO

The balanced chemical reaction is,

2Pb(s)+O_2(g)\rightarrow 2PbO(s)

From the reaction, we conclude that

As, 2 mole of Pb react to give 2 mole of PbO

So, 2.18 mole of Pb react to give 2.18 mole of PbO

Now we have to calculate the mass of PbO

\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO

\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g

Theoretical yield of PbO = 486.1 g

Experimental yield of PbO = 367.5 g

Now we have to calculate the percent yield of the reaction.

\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100

\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%

Therefore, the percent yield of the reaction is, 75.6 %

3 0
3 years ago
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