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3 years ago

Which statement is true about a reversible reaction?

Chemistry

1 answer:

bonufazy [111]3 years ago

4 0

Answer:

The correct statement is : It results in an equilibrium mixture of reactants and products.

Explanation:

A reversible chemical reaction is one that occurs in both directions, that is, of the reagents to generate products and vice versa. In synthesis, they are incomplete reactions since the reagents are not consumed in their entirety, since the products react to generate them again.

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erica [24]

Correct answer is C

Na2O + H2O ----> 2 NaOH

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4 years ago

What does e=mc2 mean

Musya8 [376]

Energy equals mass times the speed of light squared

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3 years ago

Read 2 more answers

A pH of 2 indicates a(n) O weak acid strong acicd O weak base O strong base

Crank

Strong acid is your answer

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3 years ago

What is the formula for km/day

alukav5142 [94]

Please be rightly informed that the formula for km/day is 1 kilometer per hour (kph) = 24 kilometer per day (km/day).

<h3>Fundamental quantities</h3>

These are quantities which forms the basic unit of measurement. The units of fundamental quantities are called fundamental units.

The three major fundamental quantities are:

Length
Mass
Time

Other fundamental quantities are:

Temperature
Amount of substance
Magnetic flux
Electric current

So therefore, the formula for km/day is 1 kilometer per hour (kph) = 24 kilometer per day (km/day).

Learn more about fundamental quantities:

brainly.com/question/23036403

#SPJ1

8 0

2 years ago

Calculate the enthalpy change for the reaction: CaF2+H2SO42HF+ CaSO4 Given that enthalpy changes of formation of: AHi[CaF2] = -1

son4ous [18]

Answer:

Enthalpy change of the reaction = 58 kJ/mol

Explanation:

Enthalpy of a reaction is calculated as:

$\Delta H = \Delta H_{Product} - \Delta H_{Reaction}$

$\Delta H_{CaF_2} = -1220\;kJ mol^{-1}\\\Delta H_{H_2SO_4} = -814\;kJ mol^{-1}\\\Delta H_{HF}=-271\;kJ mol^{-1}\\\Delta H_{CaSO_4} = -1434\;kJ mol^{-1}\\$

For the given reaction,

$CaF_{2} +H_2SO_4 \rightarrow 2HF + CaSO_4$

$\Delta H = (2\times \Delta H_{HF} + \Delta H_{CaSO_4}) - (\Delta H_{CaF_2} + \Delta H_{H_2SO_4})$

$\Delta H=(2\times -271 -1434) - (-1220 - 814))\;kJ/mol$

= (-1976 + 2034) kJ/mol

= 58 kJ/mol

5 0

3 years ago