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4 years ago

Bonds between metals and nonmetals tend to be ionic, but bonds between nonmetal atoms tend to be covalent.

Chemistry

2 answers:

motikmotik4 years ago

7 0

I won't do the whole thing for you, but here are points that are essential to include in your response:

Bond type depends on differences in electronegativity.

Metals and nonmetals tend to have large differences in electronegativity.

Nonmetals tend to have smaller differences in electronegativity.

Ionic bonds form between atoms that have large electronegativity differences.

Covalent bonds form between atoms with small electronegativity differences.

Hope I helped :P

I am Lyosha [343]4 years ago

5 0

O i get it it does not stik on metal eater way

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3 years ago

What do the subscripts in the formula for ethane represent?

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3 years ago

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3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

<u>(b) a solution of 0.10 M $NH_3$ </u>

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

<u>(d) a solution of 0.10 M NaOH</u>

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

<u>(e) a $1\times 10^{-4}M$ solution of $HNO_2$ </u>

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

4 years ago

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0

3 years ago

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