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3 years ago

I need help pleaseee!!!!!!!! Fast

Chemistry

2 answers:

VikaD [51]3 years ago

6 0

Answer:

C.) 1700 Pa

Explanation:

300 k - 0.3 m^3 = 17000

AfilCa [17]3 years ago

4 0

Answer:

C is the correct answer

Explanation:

Hope this helps:)

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What is the average range of the PH scale? THANKS!!

alex41 [277]

The pH scale goes from 0-14. 0-6.9 is acidic, 7 is neutral and 7.1-14 is basic

8 0

3 years ago

Read 2 more answers

A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is

Reika [66]

Answer: The pH of the buffer is 4.61

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.70

$[\text{conjuagate base}]}$ = moles of conjugate base = 3.25 moles

$[\text{acid}]$ = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

$pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61$

Hence, the pH of the buffer is 4.61

8 0

3 years ago

has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a

Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

$A + B \rightarrow C$

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

$\Delta G = -RT ln K$

Putting the given values into the above formula as follows.

$\Delta G = -RT ln K$

$-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

x = 0.1417

Hence, at equilibrium

[A] = 0.3 - 0.1417

= 0.1583 M

[B] = 0.4 - 0.1417

= 0.2583 M

[C] = 0.1417 M

5 0

3 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0

3 years ago

A 10.0 gram sample of Fe contains how many miles of Fe

Dafna1 [17]

Answer:

1.7857 moles

Explanation:

moles=mass/Mr

10/56=1.7857

moles of iron =1.7857

4 0

2 years ago