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2 years ago

How many atoms are in lead (III) bicarbonate?

1 answer:

6 0

Answer:Lead is a relatively unreactive post-transition metal. Its weak metallic character is illustrated by its amphoteric nature; lead and lead oxides react with acids and bases, and it tends to form covalent bonds. Compounds of lead are usually found in the +2 oxidation state rather than the +4 state common with lighter members of the carbon group. Exceptions are mostly limited to organolead compounds. Like the lighter members of the group, lead tends to bond with itself; it can form chains and polyhedral structures.

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Why boiling point is more when vanderwall force if more?

jek_recluse [69]

Answer:

It is the intermolecular forces acting between the molecules that cause attractions between them making them liquids or solids. The strength of Van der Waals forces depends primarily on the number of electrons in total in the molecule, so larger molecules will have higher boiling points.

Explanation:

6 0

2 years ago

Photochemical chlorination of pentane gave a mixture of three isomeric monochlorides. The principal monochloride constituted 46%

meriva

Answer:

Explanation:

In the chlorination of alkanes, the condition necessary is UV light so free radical substitution can take place. For alkanes like pentane, the primary, secondary and tertiary Hydrogen atoms (Hydrogen atoms bonded to their respective carbon) þare taken into consideration and this is because the tertiary Hydrogen is the most reactive (due to bond dissociation energy) hence the easiest to be substituted. The trend is as follows in the order of their reactivity;

1° < 2° < 3°

So, the products of the chlorination of pentane, the principal monochloride constituted is 3 - chloropentane while the other two monomers are:

2- chloropentane

1- chloropentane

Below is the attachment showing the structural formula of the three monochloride constituted pentane.

5 0

3 years ago

On the moon the acceleration due to gravity is 1.6 ms if an object has a mass on earth of 5.1kg what is its mass on the moon

Y_Kistochka [10]

Answer:

Mass on the moon is 5.1 kg.

Explanation:

Given mass on earth = 5.1 kg

acceleration due to gravity on moon = 1.6 ms⁻².

The mass on the moon does not change because mass is the quantity of matter in the body regardless of its volume or of any forces acting on it.

Whereas the weight of the object changed.

Weight is the the force exerted on a body which is related to mass and expressed as W = mg.

W = weight of object m = mass of object and g = gravitational force

$So\ in\ this\ case\ W\ =1.6\times\ 5.1\ =8.16N$

N stands for newton unit of weight.

Thus mass of the object on moon is same as on earth which is 5.1 kg.

3 0

2 years ago

If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L

Tom [10]

Answer:

Approximately $2.53\; \rm L$ (rounded to three significant figures) assuming that ${\rm HCl}\, (aq)$ is in excess.

Explanation:

When ${\rm HCl} \, (aq)$ and ${\rm AgNO_3}\, (aq)$ precipitate, ${\rm AgCl} \, (s)$ (the said precipitate) and $\rm HNO_3\, (aq)$ are produced:

${\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)$ (verify that this equation is indeed balanced.)

Look up the relative atomic mass of $\rm Ag$ and $\rm Cl$ on a modern periodic table:

$\rm Ag$ : $107.868$ .
$\rm Cl$ : $35.45$ .

Calculate the formula mass of the precipitate, $\rm AgCl$ :

$\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of $\rm AgCl$ formula units in $45.00\; \rm g$ of this compound:

$\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}$ .

Notice that in the balanced equation for this reaction, the coefficients of ${\rm AgNO_3} \, (aq)$ and ${\rm AgCl}\, (s)$ are both one.

In other words, if ${\rm HCl}\, (aq)$ (the other reactant) is in excess, it would take exactly $1\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $1\; \rm mol \!$ of ${\rm AgCl}\, (s)\!$ formula units.

Hence, it would take $0.313987\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $0.313987\; \rm mol\!$ of ${\rm AgCl}\, (s)\!$ formula units.

Calculate the volume of the ${\rm AgNO_3} \, (aq)\!$ solution given that the concentration of the solution is $0.124\; \rm mol \cdot L^{-1}$ :

$\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}$ .

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of ${\rm AgNO_3} \, (aq)\!$ .)

In other words, approximately $2.53\; \rm L$ of that ${\rm AgNO_3} \, (aq)\!$ solution would be required.

4 0

2 years ago

Hudson is already 40 miles away from home on his drive back to college. He is driving 65 mi/h. Write an equation that models the

slega [8]

D = 65h + 40
where h is the # of hours and d is the total distance

3 0

3 years ago