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madam [21]
3 years ago
11

Given the setup of the equilibrium constant expression, a reaction that has a high equilibrium constant will "favor" the product

s
True
False
Chemistry
1 answer:
ss7ja [257]3 years ago
5 0

Answer:

for points

Explanation:

loll

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To do an experiment with the peptide H-A-P-P-Y you need to make two buffers. One where the peptide has a net negative charge (Bu
nata0808 [166]

Answer:

Buffer A: pH 5.25; 2.82 mol of acetate for every 1 mol of acetic acid

Buffer B: pH 9.25; 11.2 mol of bicarbonate for every 1 mol of carbonate

Explanation:

The pI of a protein is the pH at which it has a net charge of zero.

The protein has a positive charge below the pI and a negative charge above it.

Since pI = 7.25, let's make one buffer at pH = 5.25 and one at pH 9.25.

1. Preparation of pH 5.25 buffer

(a) Choose the conjugate pair

We should choose a buffer with pKₐ close to 5.25.

Acetic acid has pKₐ = 4.8, so let's make acetate buffer.

(b) Protocol for preparation

We can use the Henderson-Hasselbalch equation to get the acid/base ratio.

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\5.25& = & 4.8 +\log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\0.45& = & \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\2.82 & = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\end{array}\\\text{The base/acid ratio must be $\mathbf{2.82:1}$}

To make the buffer, mix the solutions to get a ratio of 2.82 mol of acetate for every 1 mol of acetic acid.

2. Preparation of pH 9.25 buffer

(a) Choose the conjugate pair

We should choose a buffer with pKₐ close to 9.25.

Bicarbonate has pKₐ = 10.3, so let's make a pH 9.25 bicarbonate buffer.

(b) Protocol for preparation

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\9.25& = & 10.3 +\log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\-1.05& = & \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\0.08913& = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\dfrac{1}{11.2}& = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\end{array}\\\text{The acid/base ratio must be $\mathbf{11.2:1}$}

To make the buffer, mix the solutions to get a ratio of 11.2 mol of bicarbonate for every 1 mol of carbonate.

3 0
3 years ago
The number of valence electrons for any element can be determined by knowing
strojnjashka [21]
The electronic configuration [the valence electrons is known by the last no after
8 0
3 years ago
The combustion reaction of propane is described by the reaction. C3H8 + 5O2 ? 4H2O + 3CO2. How many moles of O2 are required to
Minchanka [31]

the Answer would be one mole

7 0
3 years ago
If the temperature of air in gabor's lungs is 37∘c (98.6∘f), and the volume is 6l, how many moles of air n must be released by t
umka21 [38]
Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.

T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
Δn = 0,589 mol - 0,2356 mol = 0,3534 mol.
5 0
4 years ago
A rectangular tank is filled with water to a depth = h, the tank width and length are w and 1 respectively. Calculate the ratio
AnnZ [28]

Answer:

\frac{\textup{2}}{h}

Explanation:

Given:

Depth = h

Width = w

Length  = 1

Now,

Let ρ be the density of water

Thus,

Pressure force at the bottom will be

= Pressure × Area

= ρgh × (w × 1)

also,

Pressure force at the sides will be = \frac{\textup{1}}{\textup{2}}\times ρgh × (h × w)

Therefore,

The ratio of the pressure force at the bottom to the pressure force on the side of the tank will be

= \frac{\rho gh\times w\times1}{\frac{1}{2}\times\rho gh\times(h\timesw)}

= \frac{\textup{2}}{h}

8 0
3 years ago
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