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PolarNik [594]
3 years ago
9

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 

Chemistry
1 answer:
NARA [144]3 years ago
5 0

Answer:

ΔH = q_{p}

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

q_{com} = q_{p}

where

q_{p} = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

q_{com}  = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is q_{p}

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = q_{p} + 0

==> ΔH = q_{p}

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How much heat is required to vaporize 25g of water at 100*c
ElenaW [278]

Answer:

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals ,

Q = mL,

Where, m is the mass of water present

           L = specific latent heat of vaporization

Here , m= 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25×533 = 13,325 Calories

Or , Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.

4 0
3 years ago
How many liters of C3H6O are present in a sample weighing 25.6 grams?
Romashka [77]

Answer:

V = 0.0327 L.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the liters of C3H6O by the definition of density. We can tell the density of this substance as that of acetone (0.784 g/mL) and therefore calculate the liters as shown below:

V=25.6g*\frac{1mL}{0.784g}*\frac{1L}{1000mL}\\\\V=0.0327L

Regards!

7 0
3 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
Explain the digestive process in an essay.​
fiasKO [112]

Refer to attachment for your answer

8 0
2 years ago
10.
Andru [333]

Answer: 1.8 moles Fe and 2.7 moles CO_2 are produced.

Explanation:

The balanced chemical reaction is:

Fe_2O_3+3CO\rightarrow 2Fe+3CO_2  

According to stoichiometry :

3 moles of CO require 1 mole of Fe_2O_3

Thus 2.7 moles of CO will require=\frac{1}{3}\times 2.7=0.9moles  of Fe_2O_3

Thus CO is the limiting reagent as it limits the formation of product and Fe_2O_3 is the excess reagent.

As 3 moles of CO give = 2 moles of Fe  and 3 moles of CO_2

Thus 2.7 moles of CO will give =\frac{2}{3}\times 2.7=1.8moles  of Fe  and \frac{3}{3}\times 2.7=2.7moles  of CO_2

Thus 1.8 moles Fe and 2.7 moles CO_2 are produced.

7 0
3 years ago
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