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salantis [7]
3 years ago
13

A gas with a volume of 5.0 L at a pressure of 215 kPa is allowed to expand to a volume of 14.0 L. What is the pressure in the co

ntainer if the temperature remains constant?
Chemistry
1 answer:
jarptica [38.1K]3 years ago
8 0
Wouldnt the pressure be the kPa? if so i probably know the answer

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enyata [817]
I think it’s B 5.54 x 10^2g
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I NEED HELP PLEASE, THANKS!
ira [324]

Answer:

About 0.652

Explanation:

Because the reaction is balanced, we can go straight to the next step. The molar mass of potassium is about 39.098, while the molar mass of hydrogen gas is 2 and the molar mass of water is 18. Therefore, 25.5g of potassium would be about 0.652 moles, and 220 grams of water would be about 12.222 moles, making potassium the limiting reactant. Since there is a single unit of each compound on both sides of the equation, there would be an equal amount of moles of potassium and hydrogen, and therefore about 0.652 moles of hydrogen gas would be produced. Hope this helps!

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10 g of a solute is added to 1.00 L of pure water. Half of the added solute fails to dissolve. What can you conclude?
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If in the solution, half of the added solute fails to dissolve. The solution started out supersaturated. The correct option is b.

<h3>What is supersaturation?</h3>

Supersaturation is the condition where the solutes exceed the amount that can be dissolved in a solution.

Supersaturation occurs when the solute no longer mix in the solution.

Thus, the correct option is b. The solution started out supersaturated.

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How are carrot, an amoeba, and a bacterium alike
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Why is the answer C for this problem?
pshichka [43]

Answer:

\boxed{\text{(C) X}$_{3}$P$_{2}}

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}

We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.

We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.

So, X is in Group 2 and P is in Group 15.

Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is \boxed{\text{X}$_{3}$P$_{2}}$}.

4 0
3 years ago
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