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tamaranim1 [39]

2 years ago

8

iron combines with 4.00 g of copper (11) nitrate to form 6.01 g of Iron (l) nitrate and 0.400 g copper metal. how much iron did

it take to convert to Cu(NO3)2?

1 answer:

kvasek [131]2 years ago

7 0

Answer:

idkkk

Explanation:

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Which of the following is not a nanoscale object?

KiRa [710]

If I am correct I would believe that it would be a muscle cell.

6 0

1 year ago

Read 2 more answers

The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V 0 V0 for an enzyme‑

ipn [44]

Answer:

The Michaelis‑Menten equation is given as

v₀ = Kcat X [E₀] X [S] / (Km + [S])

where,

Kcat is the experimental rate constant of the reaction; [s] is the substrate concentration and

Km is the Michaelis‑Menten constant.

Explanation:

See attached image for a detailed explanation

3 0

2 years ago

2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:

Nesterboy [21]

Answer

increase in temperature

decrease in pressure

continuous removal of PH3
adding more of P into the system

Explanation:

In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);

The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:

heat+A⇌BΔH=+

Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.

According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.

4 0

1 year ago

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(c) $Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

6 0

2 years ago

Arranges the following molecules in order of increasing dipole moment: <br> H2O, H2S, H2Te, H2Se.

erastova [34]

Explanation:

Dipole moment is defined as the measurement of the separation of two opposite electrical charges.

$H_{2}O$ is a bent shaped molecule with a dipole moment of 1.87.

$H_{2}S$ is also a bent shaped molecule with a dipole moment of 1.10.

$H_{2}Te$ is a also a bent shaped molecule and has a negligible dipole moment.

$H_{2}Se$ has a dipole moment of 0.29.

Therefore, given molecules are arranged according to their increasing dipole moment as follows.

$H_{2}Te$ < $H_{2}Se$ < $H_{2}S$ < $H_{2}O$

7 0

2 years ago