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4 years ago

Which phase change at standard temperature and pressure represents sublimation? [ASSET]

Chemistry

1 answer:

Alexus [3.1K]4 years ago

7 0

(a) CO₂(s) ⇌ CO₂(g) is an example of sublimation.

<em>Sublimation</em> is the conversion of a <em>solid directly to a gas</em> without going through the liquid phase.

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At 189k a sample of gas has a volume of 32.0cm. what does the gas occupy at 242k

sattari [20]

<h2>Answer:</h2>

At 189 K volume = 32.0 cm

As volume of gas is directly proportional to temperature.

As the temperature increase the volume of gas will also increase.

Volume at 1 K = 32/189

Volume at 242 K = 32/189 * 242 = 40.97 cm.

Gas occupy 40.97 cm volume at 242 K.

5 0

4 years ago

Contrast the way the conqueror comes with the way the Pilgrims came (stanzas three and four).

Masja [62]

Answer:

tbh id k i've never read it so if u could send the link I could help you.

Explanation:

7 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:

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

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4 years ago

How many moles are in 9 grams of water

shtirl [24]

9g x 1/18.02= 0.499 rounded is 0.50

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3 years ago

What is the wavelength of a 1.28 x 10^17 hz wave

taurus [48]

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3 years ago