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serg [7]
3 years ago
7

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of carbonic acid and sodium hydroxide are

combined. Assume excess base.
Chemistry
1 answer:
goldenfox [79]3 years ago
3 0

Answer:

H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up this net ionic equation, by firstly setting up the complete molecular equation as follows:

H_2CO_3(aq)+2NaOH(aq)\rightarrow Na_2CO_3(aq)+2H_2O(l)

Thus, since carbonic acid is weak it merely ionizes whereas sodium hydroxides ionizes for the 100 % as it is strong; thus, we can write the complete ionic equation:

H_2CO_3(aq)+2Na^+(aq)+2OH^-(aq)\rightarrow 2Na^+(aq)+(CO_3)^{2-}(aq)+2H_2O(l)

Whereas sodium ions act as the spectator ones to be cancelled out for us to obtain:

H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)

Regards!

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If 135.7 J of heat are added to 54.0 g of water initially at 25.0 °C . What is the final temperature of the water?
german

Answer:

The final temperature of the water is 25.6 °C.

Explanation:

We have,

Heat added to water is 135.7 J

Mass, m = 54 g

Initial temperature was 25 °C

It is required to find the final temperature of the water. The heat added when temperature is increased is given in terms of specific heat capacity as :

Q=mc(T_f-T_i)

T_f is final temperature

c is specific heat capacity, for water, c=4.184\ J/g^{\circ} C

So,

T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{135.7}{54\times 4.184}+25\\\\T_f=25.6^{\circ} C

So, the final temperature of the water is 25.6 °C.

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4 years ago
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Kaka ml nyu yan wala kaming paki kung ano ang lagi ginagawa nyu

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Determine which equations you would use to solve the following problem: Calculate the amount of heat needed to change 20.0 g of
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Answer:

Q = 4019.4 J

Explanation:

Given data:

Mass of ice = 20.0 g

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m = mass of given substance

c = specific heat capacity of substance

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ΔT = T2 - T1

ΔT =  89.0°C - (-10°C)

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Q = 20.0 g ×2.03 J/g.°C × 99°C

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