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TiliK225 [7]
3 years ago
14

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

Chemistry
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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A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
Al2(SO4)3+Mg(NO3)2⟶ What would be the product(s) of this reaction? *These are NOT balanced, just look for the correct products*
Readme [11.4K]

Answer:

Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4

Explanation:

Hello there!

In this case, for the given reactants side, we infer this is a double replacement reaction because all the cations and anions are switched around as a result of the chemical change, we infer that the products side include aluminum with nitrate and magnesium with sulfate as shown below:

Al_2(SO_4)_3+Mg(NO_3)_2\rightarrow Al(NO_3)_3+MgSO_4

However, we need to balance since unequal number of atoms are present at both sides, thus, we do that as shown below:

Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4

Thus, we make 6 Al atoms, 3 S atoms, 3 Mg atoms and 30 O atoms on each side in agreement with the law of conservation of mass.

Regards!

7 0
2 years ago
The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.
Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions {\text{IO}_{3}}^{-}.

How many moles of iodine \text{I}_2 will be produced?

{\text{IO}_{3}}^{-} converts to \text{I}_2 in the first reaction. The coefficient in front of \text{I}_2 is three times the coefficient in front of {\text{IO}_{3}}^{-}. In other words, each mole of {\text{IO}_{3}}^{-} will produce three moles of \text{I}_2. 0.0020 moles of {\text{IO}_{3}}^{-} will convert to 0.0060 moles of \text{I}_2.

How many moles of thiosulfate ions {\text{S}_2\text{O}_3}^{2-} are required?

\text{I}_2 reacts with {\text{S}_2\text{O}_3}^{2-} in the second reaction. The coefficient in front of \text{I}_2 is twice the coefficient in front of {\text{S}_2\text{O}_3}^{2-}. How many moles of {\text{S}_2\text{O}_3}^{2-} does each mole of \text{I}_2 consume? Two. 0.0060 moles of \text{I}_2 will be produced. As a result, 2 \times 0.0060 = 0.0120 moles of {\text{S}_2\text{O}_3}^{2-} will be needed.

6 0
3 years ago
An atom has a mass number of 30 and 16 neutrons. What is the atomic number of this atom?
Lana71 [14]
The mass number is the number of protons plus the number of neutrons. Since there are 16 neutrons, there are 14 protons. This also corresponds to the atomic number, so this atom's atomic number is 14 which is also Silicon
8 0
3 years ago
Read 2 more answers
Can you determine the molecular formula of a substance from its percent composition
Natalka [10]
The percent composition<span> gives you only the empirical formula. 
</span><span>To get the molecular formula, you must either know the molecular mass or do an experiment to find it.</span>
8 0
3 years ago
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