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3 years ago

1.0 x 10^-4 M Solution of HNO3 has been prepared for a laboratory experiment. Calculate the [H+] of this solution.

1 answer:

7 0

To determine the concentration of [H+] ions in the solution, it is important that we know the dissociation reaction of the solute. For this case, it would be

HNO3 = H+ + NO3-

Therefore, there is 1:1 ratio between HNO3 and H+ ions. The concentration of the ions would be the same as the solute which is 1.0x10^-4 M.

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¿Cuál es la diferencia entre electrones internos y electrones externos? *

Fudgin [204]

Answer & Explanation:

Los electrones externos se encuentran más lejos del núcleo. El número de electrones en la capa más externa (electrones de valencia) de un átomo en particular determina su reactividad (tendencia) a formar enlaces químicos con otros átomos.

Los electrones internos son los más cercanos al núcleo. Protegen los electrones de valencia del núcleo, reduciendo la carga nuclear efectiva.

3 0

3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

brainly.com/question/2082986?referrer=searchResults

brainly.com/question/1657608?referrer=searchResults

I hope it helps you!

6 0

3 years ago

At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubil

8090 [49]

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

$\frac{S_{1}}{P_{1}}=\frac{S_{2}}{P_{2}}$

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

$\frac{0.725}{3.08}=\frac{S_{2}}{8}$

On rearranging,

$S_{2}=\frac{0.725\times 8}{3.08}=1.88$

Therefore, solubility will be 1.88 mg of $N_{2}$ gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.

3 0

3 years ago

What net ionic equation describes the reaction when these solutions are mixed?

almond37 [142]

I think the correct answer from the choices listed above is option D. The net ionic equation that will describe the reaction of the reactants when mixed is expressed as:

2PO43–(aq) + 3Ca2+(aq) → Ca3(PO4)2(s)

Hope this answers the question. Have a nice day.

7 0

4 years ago

Read 2 more answers

4.81*10^24 atoms of lithium

enyata [817]

Answer:

Mass = 55.52 g

Explanation:

Given data:

Number of atoms of Li = 4.81×10²⁴ atom

Number of grams = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For Li:

4.81×10²⁴ atom × 1 mol / 6.022 × 10²³ atom

8 moles

Mass in gram:

Mass = number of moles × molar mass

Mass = 8 mol × 6.94 g/mol

Mass = 55.52 g

4 0

3 years ago