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3 years ago

Compare and contrast the uses of the iron triad and the use of the coinage metals

1 answer:

7 0

<span>he iron triad has the most variable oxidation states from +6, etc. The coinage metals have oxidation states of +1 except for copper which can also be +2. The zinc group have common oxidation states of +2 except for mercury which can also have +1. The coinage and zinc groups cannot have oxidation states greater than 2 under normal conditions.</span>

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What's the charge on Ni in the compound NiCi3? *

ss7ja [257]

Answer:

Explanation:

Chlorine is anion with a -1 charge. But they are three chlorine atoms.

-1 * 3 = -3

So they have a -3 charge.

So to balance the compound, the nickel has to be a cation with a +3 charge.

-3 + 3 = 0

Furthermore, a chemical bond always has a 0 charge. Remember that.

Hope it helped! Rate my answer a 5 star if correct.

8 0

2 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

How does concentration affect the chemical equilibrium?

nevsk [136]

When the concentration of a reactant is increased, the chemical equilibrium will shift towards the products. More product is formed and the concentration of the reactants decreases as the concentration of the products increases.

7 0

3 years ago

When the white part is on the left the moon is.??

Kazeer [188]

Answer:

Last Quarter also called Third Quarter.

Explanation:

6 0

2 years ago

What volume of a 0.550 M solution of potassium hydroxide (KOH) can be made with 19.9 g of potassium hydroxide?

Natalija [7]

Answer:

0.645 L

Explanation:

To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.

(Step 1)

Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (KOH): 56.104 g/mol

19.9 grams KOH 1 mole
-------------------------- x ----------------------- = 0.355 moles KOH
56.014 grams

(Step 2)

Molarity = moles / volume <----- Molarity ratio

0.550 M = 0.355 moles / volume <----- Insert values

(0.550 M) x volume = 0.355 moles <----- Multiply both sides by volume

volume = 0.645 L <----- Divide both sides by 0.550

6 0

10 months ago