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3 years ago

What is hydroxide ion concentration, [OH-], in a 1.37 x 10^-5 M solution of barium hydroxide (Ba(OH)2)?

Chemistry

1 answer:

Olegator [25]3 years ago

7 0

Ba(OH)₂ is a strong base, therefore completely ionises in the solution as follows
Ba(OH)₂ --> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂ gives out 2 mol of OH⁻
therefore number of OH⁻ moles are twice the amount of Ba(OH)₂ moles in the solution
hence,
[OH⁻] = 2[Ba(OH)₂]
therefore hydroxide ion concentration = 2 x 1.37 x 10⁻⁵ M
[OH⁻] = 2.74 x 10⁻⁵ M
coefficient = 2.74
exponent = -5

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Universal indicator solution

Lynna [10]

Shred red cabbage ~ (3/4 of a very small head)

Put the cabbage pieces in a small container ~ ( you can use a Pyrex-4-cup measure, a bowl or even a plastic zipper bag)

Cover the cabbage with very hot water. Let it sleep until the water has cooled. (somewhere between lukewarm and room-temperature)

The purple liquid you've made is your indicator.
Pour it into a container and compost the cabbage.

Now look for substances that may be acids or bases.
Liquids are good, like fruits.

You can also use solids around for baking are good too. (such as baking soda, salt, sugar, cream of tartar...)

Get containers for mixing (such as tea cups, because they are small, shallow and white inside)

Pour the indicator into the tea cups and add an acid or base.

Lemon juice, rice wine vinegar, and apple cider vinegar, turn the cabbage-water indicator into a pink.

Orange juice or fresh oranges (same thing) turn the cabbage-water indicator into an orangish-pinkish color.

Baking soda turns the cabbage-water indicator blue.

Milk (non-fat) turns the cabbage-water indicator turn opaque and milky, yet purple.

An egg white (which won't get into the solution immediately until after a lot of stirring) turns the cabbage-water indicator blue.

Hint:
Bases mostly turn the indicator towards blue-ish colors such as purple, light blue, dark blue, opaque blue...

Acids mostly turn the indicator towards pink-ish colours such as orange-ish pink, floral pink...

(You'll have to keep on testing the cabbage-water indicator in after a day or two to see if the indicator quality persists or degrades.

6 0

3 years ago

Read 2 more answers

A 251 ml sample of 0.45M HCl is added to 455 mL of distilled water. What is the molarity of the

Oksi-84 [34.3K]

We are given:

251 mL sample of 0.45M HCl added to 455 mL distilled water

Whack a mole! (finding the number of moles):

We know that in order to find molarity, we use the formula:

Molarity = number of moles / Volume (in L)

so, number of moles is:

Number of moles = Molarity * Volume(in L)

now let's plug the values for the HCl solution to find the number of moles

Number of moles = 0.45M * 0.251 L

Number of moles = 0.113 moles

Time to concentrate (finding the final concentration):

Total final volume = 251 mL + 455 mL = 706 mL = 0.706 L

Number of moles of HCl = 0.113 moles

Molarity = Number of moles / Volume (in L)

Molarity = 0.113 / 0.706

Molarity = 0.16 M

___________________________________________________________

BONUS METHOD TIME!!!

We know the relation:

M1 * V1 = M2 * V2

where M1 and M2 are the initial and final molarities and V1 and V2 are initial and final volumes respectively

notice that I didn't mention that the volume has to be in Liters, that's because of the units being concerned with both sides of the equation, say I have the volume in mL and want to convert both these volumes to L, I would divide both sides by 1000, which would NOT change the overall value

Now, plugging values in this equation

(0.45) * (251) = (251 + 455)* (M2)

112.95 = (706)(M2)

M2 = 112.97/706 [dividing both sides by 706]

M2 = 0.16 Molar

8 0

3 years ago

If 11.9 kJ are used to heat a sample of water the temperature increases from 20.0°C to

Kipish [7]

Answer:

$m=4.51g$

Explanation:

Hello!

In this case, since the energy involved during a heating process is shown below:

$Q=mCp\Delta T$

Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:

$m=\frac{Q}{Cp\Delta T}$

Thus, by plugging in, we obtain:

$m=\frac{11900J}{4.184\frac{J}{g\°C}(650\°C-20.0\°C)}\\\\m=4.51g$

Best regards!

7 0

3 years ago

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

The void fraction of the bed is 0.37.

6 0

3 years ago

Which development was a direct result of the

gladu [14]

"(2) increase in suburbanization" was a direct result of the baby boom that followed World War II, but of course there were other factors that resulted as well.

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3 years ago