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2 years ago

Molecules are not reacting using current methods. Which change can encourage a reaction?

Chemistry

1 answer:

matrenka [14]2 years ago

7 0

Adding heat to a system can encourage or increases a chemical reaction.

<h3>Change that can encourage a reaction</h3>

Increasing the temperature of a process, increase occurs in the average speed of the reactant molecules and we know that when more molecules move faster, the more number of molecules reacts with other molecules which results in faster formation of products.

In conclusion, adding heat is the right option.

Learn more about reaction here: brainly.com/question/26018275

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defon

Answer:

solar eclipse

Explanation:

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3 years ago

Read 2 more answers

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

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3 years ago

Draw the products of the complete hydrolysis of an acetal. Draw all products of the reaction.

jolli1 [7]

Answer: the product is ketone or aldehyde

Explanation:

The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using

H3O+/ ROH...

Attached is the structural conversion

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Medium in size is your answer

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2 years ago

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