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I am Lyosha [343]
3 years ago
8

The zinc in a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so th

at the hydrochloric acid can get to the zinc). The reaction between the acid and the zinc is: 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq).
When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 ∘C was 0.961 L at a total pressure of 748 mmHg .
Chemistry
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

A penny dissolves in hydrochloric acid if the copper coating is filed down in several spot... ... When The Zinc In A Certain Penny Dissolves, The Total Volume Of Gas ... is filed down in several spots (so that the hydrochloric acid can reach the zinc). The reaction between the acid and the zinc 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq) .

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A cool, yellow-orange flame is used to heat the crucible. Would this affect the mass of the crucible? If so, how?
s2008m [1.1K]

Answer:

yes

Explanation:

Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.

5 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

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2 years ago
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7 0
3 years ago
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Okay
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3 years ago
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VikaD [51]

Answer:

Which question?

Explanation:

3 0
3 years ago
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