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4 years ago

Consider the following chemical reaction:

Chemistry

2 answers:

VashaNatasha [74]4 years ago

7 0

Here's your solution. Hope it helps.

Viefleur [7K]4 years ago

3 0

According to the equation, the ratio of $HC\ell$ to $C\ell_2$ is $4:1$ . Then:

$\dfrac{n_{HC\ell}}{n_{C\ell_2}}=\dfrac{4}{1}\Longrightarrow\dfrac{0.268}{n_{C\ell_2}}=\dfrac{4}{1}\Longrightarrow n_{C\ell_2}=\dfrac{0.268}{4}\iff n_{C\ell_2}=0.067~mol$

Now we can calculate the mass of $C\ell_2$ formed.

$m_{C\ell_2}=n_{C\ell_2}\cdot MM_{C\ell_2}\Longrightarrow m_{C\ell_2}=0.067\cdot(2\cdot35.5)\Longrightarrow\\\\
m_{C\ell_2}=0.067\cdot71\Longrightarrow\boxed{m_{C\ell_2}=4.757~g}$

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julsineya [31]

The answer to your question is D. Sound waves travel faster in more dense material, and the more dense the material the louder the sound is.

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3 years ago

Read 2 more answers

A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of so

Oksi-84 [34.3K]

Answer:

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

$V_{equivalence}_{acid}$ = 1.00 L

Equation for the reaction can be represented as:

$NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)$

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = $\frac{18.08g}{0.140 mole}$

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = $10^{-4.63]$

= $2.3*10^{-5}$

$[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}$

$= \frac{0.140 mol}{1.116 L}$

= 0.1255 M

$K_a$ of HA = $2.3*10^{-5}$

$K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}$

$= 4.35*10^{-10}$

$A_{(aq)}$ + $H_2O_{(l)}$ $\rightleftharpoons$ $HA_{(aq)}$ + $OH^-_{(aq)}$

Initial 0.1255 0 0

Change - x + x + x

Equilibrium 0.1255 - x x x

$K_b = \frac{[HA][OH^-]}{[A^-]}$

$4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}$

As $K_b$ is very small, (o.1255 - x) = 0.1255

$x = \sqrt{0.1255*4.35*10^{-10}}$

[OH⁻] = $x = 7.4 *10^{-6}$

But pOH = - log [OH⁻]

= - log [ $7.4*10^{-6}$ ]

= 5.13

pH = 14.00 = 5.13

pH = 8.87

6 0

3 years ago

84.9 g of solid iron reacts with oxygen gas forming iron(III) oxide. How many moles of oxygen will react

Ivenika [448]

Answer:

1.14 moles of oxygen will react

Explanation:

The balanced equation of reaction between iron and oxygen is

4 Fe + 3 O₂ → 2 Fe₂O₃

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Fe: 4 moles
O₂: 3 moles
Fe₂O₃: 2 moles

The molar mass of iron being Fe 55.85 g/mole, then the following amount of mass reacts by stoichiometry: 4 moles* 55.85 g/mole=223.4 g

You can apply the following rule of three: if by stoichiometry 223.4 grams of Fe react with 3 moles of O₂, 84.9 grams of Fe react with how many moles of O₂?

$moles of O_{2} =\frac{84.9 grams of Fe*3 moles ofO_{2} }{223.4 grams of Fe}$

moles of O₂= 1.14 moles

1.14 moles of oxygen will react

8 0

3 years ago

Consider the following reaction.

Deffense [45]

The pressure of NO2 and N2O4 would be 1.756 atm and 0.1542 atm respectively.

According to Boyle's law, the pressure of a gas is inversely proportional to its volume at a constant temperature.

Mathematically: P1V1=P2V2

When the volume of a gas is halved at a constant temperature, the pressure is doubled. If it is the volume is doubled, the pressure would be halved.

In this case, the volume of the container is halved. Meaning that the pressure of each of the gas would be doubled by the time the equilibrium is reestablished.

Thus, NO2 would double from 0.878 atm to 1.756 atm while N2O4 will double from 0.0771 atm to 0.1542 atm

More on gas laws can be found here: brainly.com/question/1190311

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3 years ago

If three electrons are available to fill five empty 3d atomic orbitals, how will the electrons be distributed in the five orbita

Readme [11.4K]

Answer is: one electron in the first three orbitals, none in the fourth or fifth orbitals. All three electrons will have parallel spins (unpaired electrons).
This is Hund's rule - orbitals of the same energy are each filled with one electron before filling any with a second, these first electrons have the same spin.

5 0

3 years ago