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Alex777 [14]
1 year ago
5

A balloon is filled to a volume of 1.6 L at 278 K. The balloon is left outside overnight and the temperature has dropped to 253

K. What is the new volume?
Chemistry
1 answer:
german1 year ago
6 0

Explanation:

Charles' law gives the relationship between the volume and the temperature of the gas. Mathematically,

Volume ∝ Temperature

i.e. \dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}

We have, V₁ = 1.6 L, T₁ = 278 K, T₂ = 253, V₂=?

V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{1.6\times 253}{278}\\\\V_2=1.45\ L

So, the new volume is 1.45 L.

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(a) At what substrate concentration would an enzyme with a kcat of 30.0 s−1 and a Km of 0.0050 M operate at one-quarter of its m
Dmitrij [34]

The missing graph is in the attachment.

Answer: (a) [S] = 0.0016M

              (b) Vmax = 3V; Vmax = \frac{3V}{2}; Vmax = \frac{11V}{10}

              (c) Enzyme A: black graph; Enzyme B = red graph

Explanation: <u>Enzyme</u> is a protein-based molecule that speed up the rate of a reaction. <u><em>Enzyme</em></u><em> </em><u><em>Kinetics</em></u> studies the reaction rates of it.

The relationship between substrate and rate of reaction is determined by the <u>Michaelis-Menten</u> <u>Equation</u>:

<u />V=\frac{V_{max}[S]}{K_{M}+[S]}<u />

in which:

V is initial velocity of reaction

Vmax is maximum rate of reaction when enzyme's active sites are saturated;

[S] is substrate concentration;

Km is measure of affinity between enzyme and its substrate;

(a) To determine concentration:

0.25V_{max}=\frac{V_{max}[S]}{0.005+[S]}<u />

<u />0.25V_{max}(0.005+[S])=V_{max}[S]<u />

<u />0.00125+0.25[S]=[S]<u />

0.75[S] = 0.00125

[S] = 0.0016M

For a Km of 0.005M, substrate's concentration is 0.0016M.

(b) Still using Michaelis-Menten:

<u />V=\frac{V_{max}[S]}{K_{M}+[S]}<u />

Rearraging for Vmax:

V_{max}=\frac{V(K_{M}+[S])}{[S]}

(b-I) for [S] = 1/2Km

V_{max}=\frac{V(K_{M}+0.5K_{M})}{0.5K_{M}}

V_{max}=\frac{V(1.5K_{M})}{0.5K_{M}}

V_{max}= 3V

(b-II) for [S] = 2Km

V_{max}=\frac{V(K_{M}+2K_{M})}{2K_{M}}

V_{max}=\frac{V(3K_M)}{2K_M}

V_{max}=\frac{3V}{2}

(b-III) for [S] = 10Km

V_{max}=\frac{V(K_{M}+10K_M)}{10K_M}

V_{max}=\frac{V(11K_{M})}{10K_{M}}

V_{max}=\frac{11V}{10}

(c) Being the affinity between enzyme and substrate, the lower Km is the less substrate is needed to reach half of maximum velocity.

Km of enzyme A is 2μM and of enzyme B is 0.5μM.

Enzyme B has lower Km than enzyme A, which means the first will need a lower concnetration of substrate to reach half of Vmax.

Analyzing each plot, notice that the red-coloured graph reaches half at a lower concentration, therefore, red-coloured plot is for enzyme B, while black-coloured plot is for enzyme A

<u />

3 0
1 year ago
Sodium chloride (NaCl) is the chemical name for table salt and potassium chloride (KCl) is a common salt substitute. Using the p
Vedmedyk [2.9K]
C because ion know I just guessed
3 0
1 year ago
Mr. Smith is hyperglycemic with a blood glucose level of 300mg/ml of blood. Explain how homeostasis would regulate his glucose l
aleksley [76]

Answer:

When the animal has eaten food and the blood glucose level in the body increases. The pancreas cells in the body detects the increase in the blood glucose which leads to increase the insulin hormone.

This decreases the blood glucose level in the level. This is how the negative feedback works in the body if the level of glucose increases.

Negative feedback is the way by which the body maintains homeostasis and maintains equilibrium in the body.

6 0
1 year ago
When 2.36g of a nonvolatile solute is dissolved in 100g of solvent, the largest change in freezing point will be achieved when t
nignag [31]

Answer:

Option c → Tert-butanol

Explanation:

To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.

The formula is:

ΔT = Kf . m . i

When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.

i = Van't Hoff factor (ions particles that are dissolved in the solution)

At this case, the solute is nonvolatile, so i values 1.

ΔT = Difference between  fussion T° of pure solvent - fussion T° of solution.

T° fussion paradichlorobenzene = 56 °C

T° fussion water = 0°

T° fussion tert-butanol = 25°

Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.

3 0
2 years ago
2.1. The lithium ion in a 250,00 mL sample of mineral water was
juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

<h3>11482 ppt of Li</h3>

4 0
2 years ago
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