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adelina 88 [10]
2 years ago
15

The size of camphor kept in store house gradually decrease in size after few а days? ​

Chemistry
1 answer:
zzz [600]2 years ago
4 0

Answer:

Because it goes through the process of sublimation.

Explanation:

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The density of gold is 19.3 g/cm3. What is the volume of a 575 gram bar of pure gold?
Gnoma [55]

Answer:

The  answer to your question is: Volume = 29.79 cm³

Explanation:

Data

Density of gold = 19.3 g/cm3

Mass = 575 grams

Formula

density = mass / volume

Volume = mass/density

Volume = 575 g / 19.3 g/cm3

Volume = 29.79 cm³

3 0
3 years ago
Read 2 more answers
The acid dissociation constant ka for an unknown acid ha is 4.57 x 10^-3 what is the base dissociation constant kb for th econju
SashulF [63]

Answer:

2.19 x 10^-12.

Explanation:-

The relation between Ka and Kb for an acid and it's conjugate base is

Ka x Kb = Kw where Kw = ionic product of water.

So Kb = 10^-14 / (4.57 x 10 ^ -3)

= 2.19 x 10^-12

4 0
3 years ago
How many grams of carbon dioxide are produced when 16 g of methane and 48 g of oxygen gas combust
larisa [96]
I believe the correct answer is 11 g
6 0
2 years ago
The name of an oxyacid has the suffix -ous acid. What is the suffix of the oxyanion?
Dima020 [189]

Answer:

-ate

Explanation:

6 0
3 years ago
How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

7 0
4 years ago
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