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Alex17521 [72]
3 years ago
14

A scientist placed a sample of lithium into a container of water. The scientist observed the lithium floating and making a buzzi

ng sounf ad gas bubbles were forming arounf it. Which conclusion is BEST supporter by this observation? A) Dissolved gases were released. B) A chemical change took place. C) Water was vaporized. D) The lithium was dissolved by the water.
Chemistry
2 answers:
AlladinOne [14]3 years ago
6 0

Answer:

b

...................

Lelechka [254]3 years ago
3 0

Answer:

icjfenvu8envu8wmdic9h3ti9vnwu8d c8ywrnfuiwhcio2rhfio2n3fu82m6fg2joend81endui3ndudjefuh

Explanation:

ncs njevt hibveynvuirn2uivnwru8cnajoncuiwrnf24nvojwnvojwnfjr2nfln24f8u1elkcjo2joc2rjkf

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Help ASAP!!!!!!!!!!!!!!!!!!!!!!!!!
BartSMP [9]

Answer:

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Explanation:            

3 0
3 years ago
A black rock that produces heat
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Anything carbon in composition
4 0
3 years ago
Read 2 more answers
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
lys-0071 [83]

Answer:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

<em><u>Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.</u></em>

<em><u>Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.</u></em>

8 0
3 years ago
How does an increase in reactant concentration affect the rate of reaction?
Trava [24]
Option A is the correct option.

In terms of the collision theory, increasing the concentration of a reactant increases in the number of collisions between the reacting species per second and therefore increases the reaction rate.
7 0
2 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
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