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3 years ago

What is the OH- ion concentration, in moles per liter, of an aqueous solution with a pH of 7?

Chemistry

1 answer:

topjm [15]3 years ago

3 0

Answer:

Explanation:

the pH scale is based on the function

1x10-14

So since pH is neutral the concentration of OH and H must be equal hence the only option in which the concetration would [OH]{H}= 1x10-14 would be when both are equal so it cannot be a or d and when they are expressed in concetration so the only option is B.

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Calculate the mass of 29.6 L of fluorine gas at STP.

Lunna [17]

Answer:

Mass = 25.08 g

Explanation:

Given data:

Volume of fluorine = 29.6 L

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Mass of fluorine = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

n = PV/RT

n = 1 atm × 29.6 L / 0.0821 atm.L/ mol.K × 273.15 K

n = 1.32 mol

Mass of fluorine:

Mass = number of moles × molar mass

Mass = 1.32 mol ×19 g/mol

Mass = 25.08 g

7 0

3 years ago

What radionuclides is used in diagnosis of atherosclerosis

aliina [53]

Answer: technetium-99.

Explanation:

This is not an answer that you can look for in a book.

You need to do some research.

You can fiind the abstract of the article titled <span>Molecular imaging of atherosclerosis using a technetium-99m-labeled endothelin derivative.</span>

The conclusion of this scientific article is that labeling with the radiosotope Tc - 99 iis feasibele to visualize the effects of an experiment inducing atherosclerosis, which gives the answer to the question posted.
.

8 0

4 years ago

How many moles of O2 are needed to react<br> with 24 moles of C2H6?<br> 2C2H6 + 702 — 4CO2 + 6H20

Anna11 [10]

Explanation:

Mole ratio of Ethane to Oxygen = 2 : 7

Moles of O2 needed = 24 moles * (7/2) = 84 moles.

7 0

3 years ago

A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.9440-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

0.0133 mole = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.171g$

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%$

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

7 0

4 years ago

Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi

uysha [10]

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is $2.9^0C$