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Vilka [71]
3 years ago
12

Please help me! This is too hard!?

Chemistry
1 answer:
Annette [7]3 years ago
5 0

Answer: 59.24 atm

Explanation:

Given that:

Original Volume of gas V1 = 2.7L

Temperature T1 = 42.7°C

Convert Celsius to Kelvin

(42.7°C + 273 = 315.7K)

Pressure P1 = 684.9 torr

Final Volume V2 = 0.14 L

Final temperature T2 = 803.1°C

Convert Celsius to Kelvin

(803.1°C + 273 = 1076.1K)

Final pressure = ?

Then, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(684.9 torr x 2.7L)/315.7K = (P2 x 0.14L)/1076.1K

1849.23/315.7 = 0.14P2/1076.1

Then, cross multiply

1849.23 x 1076.1 = 315.7 x 0.14P2

1989956.403 = 44.198P2

Divide both sides by 44.198

1989956.403/44.198 = 44.198P2/44.198

45023.67 torr = P2

Now, convert the pressure in torr to atmosphere

If 760 torr = 1 atm

45023.67 torr = Z atm

Then, cross multiply

760 torr x Z = 45023.67 torr x 1 atm

Z = 45023.67 torr / 760 torr

Z = 59.24 atm

Thus, the new pressure of the gas will be 59.24 atm

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C4H6 + 2HBr = C4H8Br2

Explanation:

Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of CnH2n−2. They are unsaturated hydrocarbons.

Furthermore, Halogenation reaction of any alkyne such as but-2-yne with any of the hydro halogenated compounds such as HBr, HCl, HF or HI will give an alkane that will have the same number of carbon atoms in the starting material and in the final product.

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6 0
3 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

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Your science teacher gives you three liquids to pour into a jar. After pouring all of them into the jar, the liquids layer as se
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Answer:

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Explanation:

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A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask
leonid [27]

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

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Explanation:

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Molarity=\frac{n\times 1000}{V_s}

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n = moles of solute

V_s = volume of solution in ml

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8 0
3 years ago
What is the energy of light with a wavelength of 697 nm?
USPshnik [31]

Answer:

Red 697 nm

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Hopefully I answer yre question

Explanation:

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