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Natali5045456 [20]
3 years ago
7

A gas in a piston starts out with a volume of 156 mL, a temperature of 28.1oC, and a pressure of 1.12 atm. If it ends with a vol

ume of 312 mL and a temperature of 87.2oC, the new pressure will be?
Chemistry
1 answer:
Korolek [52]3 years ago
3 0

Answer:

idkhfj4 HD y

Explanation:

hwhx iij id enjoyably

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Draw a Lewis structure for C2H3Cl . Include all hydrogen atoms and show all unshared electron pairs. None of the atoms bears a f
Korvikt [17]

Answer:

See attached picture.

Explanation:

Hello!

In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.

In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.

Best regards!

5 0
3 years ago
If you try to place the compound on the pan over the flame, what could possibly happen? Explain your answer. [2 marks] anyone pl
Scilla [17]

Answer:

it will probably flame up or explode or maybe start boiling

8 0
3 years ago
Write the balanced chemical equation between H2SO4 and KOH in aqueous solution. This is called a neutralization reaction and wil
emmainna [20.7K]

Answer:

0.166M

Explanation:

In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:

<em>Moles H2SO4:</em>

0.650L * (0.430mol /L) = 0.2795moles H2SO4

<em>Moles KOH:</em>

0.600L * (0.240mol / L) = 0.144 moles KOH

Moles of sulfuric acid that reacts with 0.144 moles of KOH are:

0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.

And remain:

0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.

In 0.650L + 0.600L = 1.25L:

Molar concentration of sulfuric acid:

0.2075 moles of H2SO4 / 1.25L =

<h3>0.166M</h3>
7 0
3 years ago
At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz
Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P°    (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles:   n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

<h2>P(C₆H₆) = 0.2961 atm</h2>

Hope this helps

7 0
3 years ago
(06.03 MC) A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What i
oee [108]

Answer:

New volume is 25.0 mL

Explanation:

Let's assume the gas sample behaves ideally.

According to combined gas law for an ideal gas-

                         \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where, P_{1} and P_{2} represent initial and final pressure respectively

V_{1} and V_{2} represent initial and final volume respectively

T_{1} and T_{2} represent initial and final temperature (in kelvin) respectively

Here, T_{1}=T_{2}, V_{1}=50.0mL, P_{1}=20.0atm and P_{2}=40.0atm

So, V_{2}=\frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}=\frac{(20.0atm)\times (50.0mL)}{40.0mL}=25.0mL

So, the new volume is 25.0 mL

6 0
3 years ago
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