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lesya692 [45]
3 years ago
12

The first law of thermodynamics states that . How is this also a statement of the principle of conservation of energy?

Chemistry
2 answers:
Eddi Din [679]3 years ago
8 0

The answer is a change in internal energy causes work to be done and heat to flow into the system.  

<u>Explanation:</u>

  • The first law of thermodynamics is a similar version of the law of conservation of energy where the energy can neither be created nor be destroyed, it can be transformed from one form to the other.
  • It also defines that the work is done and heat flowing into the system is due to the change in internal energy. The sum of all energy including kinetic and potential energy except the displaced energy to the surrounding is known as internal energy.
  • ΔU represents the change in internal energy of the system, Q represents the net heat transferred into the system, and W represents the net work done by the system. So +ve Q adds energy to the system and =ve W takes energy from the system. Thus ΔU=Q−W.

Harman [31]3 years ago
7 0

Answer:

i believe that the answer is a

Explanation:

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Identify each of the following household substances as a strong electrolyte, weak electrolyte, or non-electrolyte: salt solution
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Answer:

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Explanation:

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Read 2 more answers
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

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