All categories

3 years ago

When the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is

2 answers:

8 0

PH = -log( [H3O+] )
so the pH is in powers of 10
1* 10^ 5 / 1*10^3 = 1* 10^2 = 100
<span> so the answer is:
D. 100 times the original content</span>

gizmo_the_mogwai [7]3 years ago

7 0

Answer : The correct option is, (D) 100 times the original content.

Explanation :

As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3

As we know that,

$pH=-\log [H_3O^+]$

The hydronium ion concentration at pH = 5.

$5=-\log [H_3O^+]$

$[H_3O^+]=1\times 10^{-5}M$ ..............(1)

The hydronium ion concentration at pH = 3.

$3=-\log [H_3O^+]$

$[H_3O^+]=1\times 10^{-3}M$ ................(2)

By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.

$\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}$

$100\times [H_3O^+]_{original}=[H_3O^+]_{final}$

From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.

Hence, the correct option is, (D) 100 times the original content.

You might be interested in

The concentration of hydrogen in soil sample 13 * 10 ^ - 6 * M the soil is tested with a pH meter, what value should the meter r

vovikov84 [41]

Answer:

pH = 4.9

Explanation:

Given data

[H⁺] = 13 × 10⁻⁶ M

The pH is a scale used to determine <em>the acidity or basicity of a solution</em>. The pH is related to the concentration of hydrogen ions through the following expression.

pH = -log [H⁺]

pH = -log 13 × 10⁻⁶

pH = 4.9

Since the pH < 7, the soil is considered to be acid.

6 0

3 years ago

Explain the ways that carbon dioxide is added to the atmosphere. How is it removed? Does most carbon enter the atmosphere as car

Licemer1 [7]

Answer:

Carbon dioxide is added to the atmosphere by human activities. When hydrocarbon fuels (i.e. wood, coal, natural gas, gasoline, and oil) are burned, carbon dioxide is released. During combustion or burning, carbon from fossil fuels combine with oxygen in the air to form carbon dioxide and water vapor.Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas. Each year, five and a half billion tons of carbon is released by burning fossil fuels.Carbon dioxide causes about 20 percent of Earth's greenhouse effect; water vapor accounts for about 50 percent; and clouds account for 25 percent.Likewise, when carbon dioxide concentrations rise, air temperatures go up, and more water vapor evaporates into the atmosphere—which then amplifies greenhouse heating

5 0

3 years ago

Nonmetals tend to boil at

Usimov [2.4K]

Answer:

D) Low temperatures

Explanation:

Non-metals are elements that do not show metallic characteristics like heat conductivity, electrical conductivity, malleability, e.t.c. They are generally less dense and are bonded with weak forces in between their compounds.

The weakness of non-metals is the chief reason why they generally tend to have low boiling point. Only the presence of impurities can cause an elevation in their boiling points.

Metals on the other hand are good heat conductors and they generally have a high boiling point.

7 0

3 years ago

Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1

Drupady [299]

Answer : The value of second ionization energy of Ca is 1010 kJ.

Explanation :

The formation of calcium oxide is,

$Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)$

$\Delta H_f^o$ = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of $CaO$ :

(1) Conversion of solid calcium into gaseous calcium atoms.

$Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)$

$\Delta H_s$ = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

$Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)$

$\Delta H_I_1$ = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

$Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)$

$\Delta H_I_2$ = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

$O_2(g)\overset{\Delta H_D}\rightarrow OI(g)$

$\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)$

$\Delta H_D$ = dissociation energy of oxygen = $\frac{498}{2}=249kJ$

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

$O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)$

$\Delta H_E_1$ = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

$O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)$

$\Delta H_E_2$ = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

$Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)$

$\Delta H_L$ = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L$

Now put all the given values in this equation, we get:

$-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)$

$\Delta H_I_2=1010kJ$

Therefore, the value of second ionization energy of Ca is 1010 kJ.

6 0

3 years ago

The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the

enot [183]

Answer:

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g; H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4 or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

C₁H₂ = CH₂

Therefore, sample A is not the same as propene.

Sample B:

C: 72 g H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

H: 12 / 1 = 12 ----> 12 / 6 = 2

EF: CH₂

Sample C:

C: 84 g H: 10 g

C: 84 / 12 = 7 ----> 7 / 7 = 1

H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g H: 10 g

C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1

H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6 -----> CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

Hope this helps

8 0

3 years ago