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3 years ago

14

If more products are present in a mixture ,what can be said of the reversible reaction?

2 answers:

motikmotik3 years ago

7 0

D. The reaction occurring in the forward direction is favored.

<span>In a reversible reaction, both the forward and reverse reactions take place (at equilibrium) </span>
<span>A. at the same rate </span>

<span>Which of the following is untrue concerning a reversible reaction? (at equilibrium) </span>
<span>D. The reaction could spend more time in the reverse direction than in the forward direction. </span>

<span>A reversible reaction </span>
<span>C. always starts off in the forward direction. </span>

<span>All reversible reactions will reach chemical _____. </span>
<span>D. equilibrium</span>

lys-0071 [83]3 years ago

7 0

Answer:

The reaction occurring in the forward direction is favored.

Explanation:

Consider the reversible reaction

A+B ----------------->. C

<------

It can easily be seen that the equilibrium position lies more to the product side than the reactants side( the longer arrow points towards the products while the sorter one points towards reactants). That is, the forward reaction (product formation) is favoured under a given set of conditions. Hence there will be more products in the reaction system than reactants.

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Is sugar a compound or a mixture?

puteri [66]

Answer:

compound

Explanation:

3 0

2 years ago

Read 2 more answers

Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of

Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

5 0

3 years ago

Which sequence repesent matter that is losing energy

astra-53 [7]

Which sequence repesent matter that is losing energy

answer: solid liquid gas

HOPE THIS HELPS :)

5 0

3 years ago

When liquid heats up is it a chemical change

Mazyrski [523]

Yes because of the gas combination.

5 0

3 years ago

Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/

ankoles [38]

The balanced chemical equation for the combustion of butane is:

$2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)$

Δ $H_{reaction}^{0}$ = Σ $n_{products}$ Δ $H_{f}^{0}_{(products)}$ -Σ $n_{reactants}$ Δ $H_{f}^{0}_{(reactants)}$

= $[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]$ =[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

= -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

$1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}$

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

6 0

3 years ago