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2 years ago

Select all the correct answers.

Chemistry

2 answers:

rjkz [21]2 years ago

7 0

Answer:

For Plato users it is A

HClO3 + LiOH → LiClO3 + H2O

Explanation:

stich3 [128]2 years ago

6 0

Answer : The products of the acid-base reaction between $HClO_3$ and LiOH are, $LiClO_3$ and $H_2O$

Explanation :

Neutralization reaction : It is a type of chemical reaction in which an acid react with a base to give salt and water as a product that means it reacts to give a neutral solution. It is also known as acid-base reaction.

The acid-base reaction will be:

$HClO_3+LiOH\rightarrow LiClO_3+H_2O$

Therefore, the products of the acid-base reaction between $HClO_3$ and LiOH are, $LiClO_3$ and $H_2O$

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The AE of a system that releases 12.4 J of heat and does 4.2 J of work on its surroundings It is_______ jA. 16.6 B. 12.4 C. 4.2

andre [41]

ANSWER

EXPLANATION

Given that

The energy released by the system is 12.4J

Work done on the surrounding is 4.2J

Follow the steps below to find the change in energy

In the given data, energy is said to be released to the surroundings

Recall, that exothermic reaction is a type of reaction in which heat is released to the surroundings. Hence, change in enthalpy is negative

Step 1; Write the formula for calculating change in energy

$\Delta E\text{ }=\text{ q }+\text{ w}$

Since heat is released to the surrounding, then q = -12J

Recall, that work done by the system on the surroundings is always negative

Hence, w = -4.2J

Step 2; Substitute the given data into the formula in step 1

$\begin{gathered} \text{ }\Delta E\text{ = q + w} \\ \text{ }\Delta E\text{ }=\text{ -12.4 }+\text{ \lparen-4.2\rparen} \\ \text{ }\Delta E\text{ = -12.4 - 4.2} \\ \text{ }\Delta E\text{ }=\text{ -16.6J} \end{gathered}$

Therefore, the change i

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