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grandymaker [24]
2 years ago
5

Fill in the products for the dissociation of the strong acid hno3 in water.

Chemistry
1 answer:
SpyIntel [72]2 years ago
3 0

The products obtained from the dissociation of HNO₃ in water are hydronium ion, H₃O⁺ and nitrate ion, NO₃¯

An acid is a substance which when dissolved in water produces hydronium ion, H₃O⁺ as the only positive ion.

Strong acid ionize completely in water while weak acid only ionize to a certain degree (i.e partially) in water

Trioxonitrate (v) acid, HNO₃ is a strong acid and will ionize complete in water to form hydronium ion, H₃O⁺ and nitrate ion, NO₃¯ as illustrated below:

HNO₃ + H₂O —> H₃O⁺ + NO₃¯

Thus, the products for the dissociation equation are: hydronium ion, H₃O⁺ and nitrate ion, NO₃¯

Learn more: brainly.com/question/13173940

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Answer:

Types of Potential Energy

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Chemical Potential Energy.

(IF THIS HELPED CAN YOU GIVE ME A BRAINYLEST PLEASE?)

3 0
3 years ago
A chemist reported that 100 gallons of a gas were available in a laboratory. Which property of the gas did the chemist report?
Cloud [144]

Answer:

It would be volume

Explanation:

6 0
3 years ago
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To what is wax susceptible
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4 0
3 years ago
When a 12.8 g sample of KCL dissolves in 75.0 g of water in a calorimeter the temp. drops from 31 Celsius to 21.6 Celsius. Calcu
Delicious77 [7]

Answer:

Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

7 0
3 years ago
~lever
aleksandr82 [10.1K]
Block and tackle d is the best answer
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