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3 years ago

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All are types of chemical reactions except __________. A) decomposition reactions B) synthetic reactions C) single replacement r

eactions

1 answer:

Morgarella [4.7K]3 years ago

8 0

B) synthetic reactions. There is no such thing as "synthetic reactions" but synthesis reactions.

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How does the A Hreaction relate to the A He of molecules involved in a reaction?

igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

<em>Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction</em>.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH <em>(1)</em>

HCl: 1/2H₂ + 1/2Cl₂ → HCl <em>(2)</em>

H₂O: H₂ + 1/2O₂ → H₂O <em>(3)</em>

NaCl: Na + 1/2Cl₂ → NaCl <em>(4)</em>

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0

3 years ago

Phase changes!a. Write out the Clapeyron equation and explain what it describes.b. Write out the Clausius-Clapeyron equation and

djverab [1.8K]

Answer:

Solutions in the attachment.

Explanation:

4 0

3 years ago

Which metals have similar chemical properties? Check all that apply. barium

rodikova [14]

Alkina metals all have the similar proteries

6 0

3 years ago

Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.

maria [59]

The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%

5 0

3 years ago

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

Eddi Din [679]

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.

8 0

3 years ago

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