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3 years ago

Differentiate between a single atom of gold and a single atom of lead. (No Multiple Choice)

1 answer:

7 0

Lead has a different chemistry than most biological metals, but it can mimic some of them, such as calcium, iron, and zinc, because it tends to form 2+ ion like other metals, and these can be formed inside the body. Due to it different chemistry, lead prevents normal functioning of the enzymes it binds to, inhibiting them and poisoning the body.

Gold rarely forms ions, unless under extreme conditions that are not present in the human body. Gold ions, when ingested, are actually very toxic, so gold poisoning is possible. But because gold are so rare, gold poisoning is also a rare phenomenon.

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3 years ago

A particular reactant decomposes with a half-life of 113 s when its initial concentration is 0.372 M. The same reactant decompos

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Answer:

Rate constant = 0.0237 M-1 s-1, Order = Second order

Explanation:

In this problem, it can be observed that as the concentration decreases, the half life increases. This means the concentration of the reactant is inversely proportional to the half life.

The order of reaction that exhibit this relationship is the second order of reaction.

In the second order of reaction, the relationship between rate constant and half life is given as;

t1/2 = 1 / k[A]o

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k = 1 / t1/2 [A]

Uisng the following values;

k = ?

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3 years ago

Ammonia (NH3) reacts with oxygen to form nitric oxide (NO) and water vapor: 4NH3 + 502 4NO + 6H2O b) When 20.0 g NH3 and 50.0 g

Solnce55 [7]

Answer: a) . Ammonia is the limiting reagent

b. Oxygen is left over and 0.1375 g of oxygen is left over.

c. The theoretical yield of $NO$ is 35.29 g.

d. The theoretical yield of $H_2O$ is 31.74 g.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For $NH_3$

Given mass of ammonia = 20.0 g

Molar mass of ammonia = 17.031 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{20.0g}{17.031g/mol}=1.17mol$

For $O_2$

Given mass of oxygen gas = 50.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{50.0g}{32g/mol}=1.6mol$

The chemical equation for the reaction is

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

By Stoichiometry of the reaction:

4 moles of ammonia reacts with = 5 moles of oxygen

So 1.17 moles of ammonia will react with = $\frac{5}{4}\times 1.17=1.4625mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.

Thus ammonia is considered as a limiting reagent because it limits the formation of product.

1. By Stoichiometry of the reaction:

4 moles of ammonia produces = 4 moles of $NO$

1.17 moles of ammonia will produce = $\frac{4}{4}\times 1.17=1.17moles$ of $NO$