Which ions have ten electrons in the outermost subshell?

Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml.

OleMash [197]

Answer: -

15.55 M

35.325 molal

Explanation: -

Let the volume of the solution be 1000 mL.

Density of nitric acid = 1.42 g/ mL

Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid

= 1000 mL x 1.42 g/ mL

= 1420 g.

Percentage of HNO₃ = 69%

Amount of HNO₃ = $\frac{69} {100} x 1420 g$

= 979.8 g

Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol

Number of moles of HNO₃ = $\frac{979.8 g}{63 g/ mol}$

= 15.55 mol

Molarity is defined as number of moles per 1000 mL

We had taken 1000 mL as volume and found it to contain 15.55 moles.

Molarity of HNO₃ = 15.55 M

Mass of water = Total mass of nitric acid solution - mass of nitric acid

= 1420 - 979.8

= 440.2 g

So we see that 440.2 g of water contains 15.55 moles of HNO₃

Molality is defined as number of moles of HNO₃ present per 1000 g of water.

Molality of HNO₃ = $\frac{15.55 x 1000}{440.2}$

= 35.325 molal

3 0

3 years ago

A 15.0 ml sample of gas at 10.0 degree Celsius and 760 torr changes to a pressure of 1252 torr at 35.0 degree Celsius. What is t

netineya [11]

Answer:

9.91 mL

Explanation:

Using the combined gas law equation as follows;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (torr)

P2 = final pressure (torr)

V1 = initial volume (mL)

V2 = final volume (mL)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

V1 = 15.0mL

V2 = ?

P1 = 760 torr

P2 = 1252 torr

T1 = 10°C = 10 + 273 = 283K

T2 = 35°C = 35 + 273 = 308K

Using P1V1/T1 = P2V2/T2

760 × 15/283 = 1252 × V2/308

11400/283 = 1252V2/308

Cross multiply

11400 × 308 = 283 × 1252V2

3511200 = 354316V2

V2 = 3511200 ÷ 354316

V2 = 9.91 mL

4 0

3 years ago

An Equilibrium with Cobalt Complex Ions.Investigate the equilibrium between two different complex ions of cobalt. The reaction i

Paladinen [302]

The change in color from blue to pink of the cobalt complexes here has been the basis of cobalt chloride indicator papers for the detection of the presence of water. It is also used in self-indicating silica gel desiccant granules.

Pink cobalt species + chloride ions ⇌ Blue cobalt species + water molecules

Explanation:

The adjustment in color from blue to the pink of the cobalt complexes here has been the premise of cobalt chloride indicator papers for the detection of the presence of water. It is likewise utilized in self-demonstrating silica gel desiccant granules.

Pink cobalt species + chloride particles ⇌ Blue cobalt species + water molecules

The response of [Co(H2O)6]2+(aq) + 4Cl–(aq) → [CoCl4]2–(aq) + 6H2O(l) is endothermic. In this manner, as per Le Chatelier's rule, when the temperature is raised, the situation of the balance will move to one side, shaping a greater amount of the blue complex particle at the expense of the pink species.

Including concentrated hydrochloric raises the chloride particle fixation, making the equilibrium move to one side, as per Le Chatelier. Including water brings down the chloride particle fixation, moving the equilibrium the other way.

As an extension, it is conceivable to show that it is the Cl–particles in the hydrochloric acid that move the balance by including a spatula of sodium chloride rather than the pink arrangement. This delivers a bluer color, however, this may take some time because the salt is delayed to dissolve.

5 0

3 years ago

What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not

Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

1,7378 = [A⁻]/[HA] (1)

As concentration of buffer is 0,15M, it is possible to write:

[A⁻] + [HA] = 0,15M (2)

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.

i hope it helps!

7 0

3 years ago

A phase diagram assumes ______ is kept constant. temperature volume pressure density

algol [13]

pressure......................................

6 0

3 years ago

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