Nitrogen is important to life because it is part of

In a second-order reaction (one that is second order in one reactant), cutting in half the concentration of that reactant will h

svlad2 [7]

A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.

4 0

3 years ago

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Complete the following single replacement reaction. If they don’t react, just write “NR”

Kipish [7]

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq) + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

4 0

3 years ago

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o

Damm [24]

Answer : The partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of $HBr$ = $1.0\times 10^{-2}atm$

The partial pressure of $H_2$ = $2.0\times 10^{-4}atm$

The partial pressure of $Br_2$ = $2.0\times 10^{-4}atm$

$K_p=4.2\times 10^{-9}$

The balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴

At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}$

Now put all the values in this expression, we get :

$4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}$

$p=-1.99\times 10^{-4}$

The partial pressure of $H_2$ at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

4 0

3 years ago

Given the reaction: 3() + () → 3() + 2()

Scilla [17]

Answer:

i think d is the answer for this

3 0

2 years ago

Part V Write the (a) half reaction of oxidation, (b) half reaction of reduction, & (c) the balanced redox reaction for all o

Eduardwww [97]

Answer:

Explanation:

a. Oxidation : 2O + 4e^- ------> 2O^2-

b. Reduction: 2Sr - 4e- -------> Sr^2+

c. Balanced redox reaction

2Sr + O2 ------------> 2Sr O

Oxidation and reduction can be defined by various means, addition of oxygen, removal of hydrogen, removal of electrons. For this reaction, this definition is used, oxidation is the loss of electrons while reduction is the gaining of electrons.

In (a) oxidation half reaction, the valency of oxygen is zero and then moves into lossing two electrons resulting into -2 valency.

In (b) reduction half reaction, the valency of Sr is zero and gains electrons resulting into valency of 2.

In the overall redox reaction, Sr and O2 with valency of 0 each reacts together and form SrO with valency of 2 and -2 respectively, which gives 0 and then balances the equation.

8 0

3 years ago