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3 years ago

Will a reaction occur in KCl +Br2= KBr + Cl2

1 answer:

4 0

Yes, a reaction will occur. Potassium chloride react with bromine to produce potassium bromide and chlorine would take place in a boiling solution.

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Ethanoic acid has the lowest vapour

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3 years ago

Calculate the percent ionization of ha in a 0.10 m solution. express your answer as a percent using two significant figures.

postnew [5]

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4 years ago

Read 2 more answers

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is

mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol$

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = $\frac{-9078.57}{2}=-4539.28kJ$

For per gram of compound:

Molar mass of $B_5H_9$ = 63.12 g/mole

$\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g$

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

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3 years ago

These are three theorized causes of A) glacial melting. B) changing of the climate. C) high and low ocean tides. D) movement of

professor190 [17]

actually its D) movement of tectonic plates

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4 years ago

Acetyl CoA turns the citric acid cycle one turn. How many turns will the citric acid cycle turn when 57.3 g of glucose (C6H1206)

lutik1710 [3]

The number of turns that citric acid cycle will turn when 57.3 g of glucose (C6H1206) is oxidized in the cells is 1.92 × 10²³ turns.

<h3>How to calculate number of turns?</h3>

According to this question, Acetyl CoA turns the citric acid cycle one turn.

First, we find the number of moles of 57.3 grams of glucose by dividing by its molar mass of 180g/mol.

no of moles = 57.3/180 = 0.32moles

Next, we multiply the number of moles by Avogadro's number i.e. 6.02 × 10²³ molecules.

no of turns = 6.02 × 10²³ × 0.32

no of turns = 1.92 × 10²³ turns

Therefore, the number of turns that citric acid cycle will turn when 57.3 g of glucose (C6H1206) is oxidized in the cells is 1.92 × 10²³ turns.

Learn more about no of turns at: brainly.com/question/1747943

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4 0

2 years ago