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3 years ago

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A fishing line sinker contains 0.650 moles of lead. How many atoms of lead are in the sinker? Show all of your work, including s

tarting quantity, conversion factor, and units.

1 answer:

34kurt3 years ago

7 0

Answer:

3.91x10²³ atoms of lead

Explanation:

In chemistry, a mole of a substance is defined as 6.022x10²³ particles that could be atoms, molecules, ions, etc.

As you can see, in the problem, you have 0.650moles of lead in a fishing line sinker, the present atoms are:

$0.650mol Pb \frac{6.022x10^{23}atoms}{1mol}$ =<em> 3.91x10²³ atoms of lead</em>

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What is the molar mass of 81.50 g of gas

garik1379 [7]

The molar mass of gas = 238.29 g/mol

<h3>Further explanation</h3>

Given

mass = 81.5 g

P=1.75 atm

V=4.92 L

T=307 K

Required

molar mass

Solution

The gas equation can be written

$\large{\boxed{\bold{PV=nRT}}$

$\tt n=\dfrac{mass}{molar~mass}$

So the equation becomes :

$\tt Molar~mass=\dfrac{mRT}{PV}$

Input the value :

$\tt M=\dfrac{81.5\times 0.082\times 307}{1.75\times 4.92}\\\\M=238.29~g/mol$

5 0

3 years ago

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Is iced tea with no ice a homogenuous or a heterogenuous mixture? ...?

yan [13]

It is a homogeneous mixture because you cannot see the individual components that make up the iced tea (such as the water, the molecules found in the tea leaves, etc.). Iced tea with ice in it is considered a heterogeneous mixture because you can distinguish the tea from the ice.

3 0

3 years ago

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

<h3>What are pH and pOH?</h3>

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

<u />

pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

<u />

The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

3 0

3 years ago

Complete and balance the following acid-base equations:

Alchen [17]

Answer:

a) $HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

b) $H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

c) $Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

Explanation:

a)

when $HClO_4$ is added to LiOH, lithium chlorate and water is formed.

$HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous $H_2SO_4$ is added to NaOH, $Na_2SO_4$ and $H_2O$ is formed. It is a neutrilization reaction.

$H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

No. of Na atom in left hand side = 1

No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

Now, it can been seen that all the atoms are balanced.

c) When $Ba(OH)2$ reacts with HF, baroum fluoride and water is formed.

$Ba(OH)2 + HF\rightarrow BaF_2 + H_2O$

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

$Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O$

In order to balance H and O on either side, multiply H2O by 2.

$Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

7 0

3 years ago

A block of metal has a mass of 0.0694 kg. It displaces 3.6 mL of water. What is it’s density in g/cm^3

Ivan

Answer:

19.28 g/cm^3 to the nearest hundredth.

Explanation:

The volume of water displaced = the volume of the metal.

density = mass / volume

0.0694 kg = 0.0694 * 1000

= 69.4 g.

Density = 69.4 / 3.6

= 19.28 g/cm^3.

4 0

3 years ago