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Lelu [443]
3 years ago
5

Which do you think would be safer for heating a flammable liquid sample contained in an open beaker, and Bunsen burner or an ele

ctric hot plate? Why?
Chemistry
1 answer:
Ray Of Light [21]3 years ago
6 0

Answer:

Electric hot plate

Explanation:

Considering that the liquid sample is flammable, it wouldn’t be a good idea to have it near a source of fire which the Bunsen burner provides, rather it would be better to heat through an electric hot plate as its not near a source of flame and instead produces heat through electricity.

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Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system,
givi [52]

Answer: The order from the Most energy released to most Energy   Absorbed Is given as  2---> 4--->,3-->---> 1 

B)-61.9 kJ

Explanation:

The change in the internal energy of a system  is positive if the reaction absorbs energy and  negative if the reaction releases energy. For a system to cause an increase in volume, it must have very high energy built up to be released.

1. Surroundings get colder and the system decreases in volume. Here, the surrounding absorbs energy  resulting in positive  ΔE

2. Surroundings get hotter and the system expands in volume.  Here energy is released causing the system to be negative

3. Surroundings get hotter and the system decreases in volume. Although there is a decreased volume, the system is negative because it releases energy

4. Surroundings get hotter and the system does not change in volume.  System is negative because it releases energy even thgoygh there is no change in volume

Therefore the order from the Most energy released to most Energy   Absorbed Is given as  2---> 4--->,3-->---> 1

b) Using  

 ΔE = q+ w     from 1st law of thermodynamics

 ΔE=  ΔH - P  ΔV

gIven  

 ΔH = -75.0KJ

volume=  A change  from 5.0L TO 2.0L = Final volume - initial volume = 2-5= -3.00L

P= 43.0atm

ΔE=  ΔH - P  ΔV

P  ΔV  = 43 atm x -3 = -129L.atm

We first convert  L-atm to Joules.

1 L-atm = 101.325 Joules.  

129L.atm = 129 x 101.325 = - 13071 J

to KJ becomes

13071/1000 = - 13.071KJ

Recall ΔE=  ΔH - P  ΔV and putting values

ΔE  = -75.0 - (-13.071 KJ)= -75.0 kJ + 13.071 kJ = -61.9 kJ

8 0
3 years ago
13. List the following ions in order of increasing radius: Li+, Mg2+, Br"
Sergeeva-Olga [200]

Answer:

The answer would be B, PC13

Explanation:

5 0
2 years ago
Read 2 more answers
The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react
murzikaleks [220]

<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

The given chemical equation follows:

                  4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

Equilibrium concentration of nitrogen gas = 3.00\times 10^{-3}M=0.003

Evaluating the value of 'x', we get:

\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M

Now, equilibrium concentration of ammonia = 0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M

Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

Equilibrium concentration of water = 6x=(6\times 0.0015)]=0.009M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

Hence, the value of K_{eq} is 4.84\times 10^{-5}

4 0
3 years ago
4AI+30 2 &gt; 2AI2O3 What id the chemical reaction?
vesna_86 [32]
This is a combination reaction.  Look at the 2 elements on left and a compound on the right.
7 0
3 years ago
What is the mass of a sample of NH3 containing 7.20 x 10^24 molecules of NH3?
leva [86]

Answer:203

Explanation: just have to timez

4 0
3 years ago
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