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2 years ago

Explain why groups 14 and 15 are better representatives of mixed groups than groups 13 and 16

Chemistry

1 answer:

Brrunno [24]2 years ago

3 0

<h3><u>Answer</u>;</h3>

Groups 14 and 15 each contain metals, nonmetals, and metalloids while Group 13 contains metals and a metalloid, and Group 16 contains metalloids and nonmetals.

<h3><u>Explanation;</u></h3>

Groups 13–16 of the periodic table contain one or more metalloids, in addition to metals, nonmetals, or both.
Unlike other groups of the periodic table, which contain elements in one class, groups 13–16 referred to as mixed groups contain elements in at least two different classes. In addition to metalloids, they also contain metals, nonmetals, or both.
<em><u>Group 14 also known as the carbon group contains carbon which is a non metal, silicon and germanium which are metalloids and tin and lead which are metals.</u></em>
<em><u>Group 15 also known as the Nitrogen group contains non metals such as oxygen, metalloid tellurium and a metal polonium.</u></em>

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B- ammonia. ammonia has a pH level of about 11, and anything higher than 7 is more basic and anything less than 7 is more acidic. 7 is considered neutral.

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2 years ago

Read 2 more answers

A. The measured pH of a 0.100 M HCl solution at 25 degrees Celsius is 1.092. From this information, calculate the activity coeff

ra1l [238]

Answer:

activity coefficient $\mathbf{\gamma =0.809}$

activity coefficient $\mathbf{\gamma = 0.791}$

The change in pH in part A = 0.092

The change in pH in part B = 0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] = $10^{-1.092}$

[a] = 0.0809 M

Recall that [a] = $\gamma$ × C

where;

$\gamma$ = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

$\gamma = \dfrac{[a]}{C}$

$\gamma = \dfrac{0.0809 \ M}{0.100 \ M}$

$\mathbf{\gamma =0.809}$

The pH of a solution of HCl and KCl = 2.102

[a] = $10^{-2.102}$

[a] = 0.00791 M

activity coefficient $\gamma = \dfrac{0.00791 \ M}{0.01 \ M}$

$\mathbf{\gamma = 0.791}$

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102

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2 years ago

What mass of solid NaOH (97.0 % by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 1

Lelechka [254]

<u>Answer:</u> The mass of 97 % of NaOH solution required is 114.33 g

<u>Explanation:</u>

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Density of 10 % solution = 1.109 g/mL

Volume of 10% solution = 1 L = 1000 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.109g/mL=\frac{\text{Mass of }10\%\text{ solution}}{1000mL}\\\\\text{Mass of }10\%\text{ solution}=1109g$

The mass of 10 % solution is 1109 g.

To calculate the mass of concentrated solution, we use the equation:

$c_1m_1=c_2m_2$

where,

$c_1\text{ and }m_1$ are the concentration and mass of concentrated solution.

$c_2\text{ and }m_2$ are the concentration and mass of diluted solution.

We are given:

$c_1=97\%\\m_1=?g\\c_2=10\%\\m_2=1109g$

Putting values in above equation, we get:

$97\times m_1=10\times 1109\\\\m_1=114.33g$

Hence, the mass of 97 % of NaOH solution required is 114.33 g

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2 years ago

What is the concentration of a solution made with 0.150 moles of KOH in 400.0 mL of solution?

otez555 [7]

Answer:

$concentration = \frac{0.15}{0.4}=0.375 mol/L$

Explanation:

Concentration: i is defined as the mole per litre.

$concentration = \frac{mole}{volume in L}$

mole=0.15

volume=400 ml=0.4 litre

$concentration = \frac{0.15}{0.4}=0.375 mol/L$

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2 years ago

Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu

Nitella [24]

Answer:

$\Delta \:H_{vap}=40383.88\ J/mol$

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,

P is the vapor pressure

ΔHvap is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

Given:

$P_1$ = 13.95 torr

$P_2$ = 144.78 torr

$T_1$ = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

$T_1$ = 298.15 K

$T_2$ = 75°C = 348.15 K

So,

$\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)$

$\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}$

$\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=40383.88\ J/mol$

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2 years ago