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3 years ago

Compared to the nonmetals in Period 2, the metals in Period 2 generally have larger

1 answer:

6 0

The answer is atomic radii; the size or radii of an atom increases from left to right, versus the ionization energies and electronegativities of atoms which increase from right to left.

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A 25.0 ml sample of 0.723 m hclo4 is titrated with a 0.27303 m koh solution. the h3o+ concentration after the addition of 66.2 m

tia_tia [17]

This doesn't need an ICE chart. Both will fully dissociate in water.

Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.

Step 1:

write out balanced equation for the reaction

HClO4+KOH ⇔ KClO4 + H2O

the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4

Step 2:

Determining the number of moles present in HClO4 and KOH

Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4

Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L

Remember:

M = moles/L so we have 0.025 L of 0.723 moles/L HClO4

Multiply the volume in L by the molar concentration to get:

0.025L x 0.723mol/L = 0.0181 moles HClO4.

Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH

Step 3:

Determine how much HClO4 remains after reacting with the KOH.

Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:

moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0

This means all of the HClO4 is used up in the reaction.

If all of the acid is fully reacted with the base, the pH will be neutral = 7.

Determine the H3O+ concentration:

pH = -log[H3O+]; [H3O+] = 10-pH = 10-7

The correct answer is 1.0x10-7.

3 0

3 years ago

Read 2 more answers

g Nitrogen in the atmosphere consists of two nitrogen atoms covalently bonded together (N2). Knowing that nitrogen is atomic num

vichka [17]

Explanation:

Since, the atomic number of nitrogen is 7 and its electronic distribution is 2, 5. So, in order to attain stability it needs to gain 3 electrons.

Hence, when it chemically combines another nitrogen atom then as both the atoms are non-metals. So, sharing of electrons will take place.

Also, there is no difference in electronegativity of two nitrogen atoms. Hence, compound formed $N_{2}$ is non-polar covalent in nature.

5 0

3 years ago

Describe what happens as ice changes to water.

morpeh [17]

Answer:

The energy of the particles increase and the molecules move more quickly.

Explanation:

The molecules are moving from a solid (barely moves, molecules close together) to a liquid (molecules slide past each other and take any shape), so molecules are moving more and have more energy

8 0

3 years ago

What are 4 properties that krypton has?

Nuetrik [128]

Atomic number36,Atomic mass83.80 g.mol -1,Density3.73 10-3 g.cm-3 at 20°C,Melting point- 157 °C,Boiling point- 153° C

5 0

3 years ago

Read 2 more answers

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago