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3 years ago

Match each of the following terms from the Michaelis-Menten equation to its correct definition.

Chemistry

1 answer:

Nadya [2.5K]3 years ago

5 0

Answer:

d. 1:B; 2:D; 3:C; 4:A

Explanation:

The Michaelis-Menten equation is:

$V_{0}=V_{max}\frac{[substrate]}{[substrate]+K_{m}}$

Also

$V_{0}=K_{cat}E_{0}\frac{[substrate]}{[substrate]+K_{m}}$

Where

Kcat = turn over number [ number of substrate molecules turned over by enzyme]

Vmax = Maximum rate (rate of a reaction when enzyme is saturated with substrate)

Km = is the substrate concentration when the V is numerically equal to half of Vmax

Kcat / Km = Enzyme efficiency.

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Fill in the blanks: Electrons are ___ in an ionic bond, whereas they are___ in a polar covalent bond, and ___ in a nonpolar cova

NNADVOKAT [17]

Answer:

The correct answer is "transferred; unequally shared; equally shared".

Explanation:

Ionic bonding occurs when a positively charged atom (cation) interacts with a negatively charged atom (anion). In ionic bonding, the cation transfers its electron to the anion. In polar covalent bonding, electrons are unequally shared. This means that the electrons spend more time in an atom than the other, which gives partial positive and negative charges to the atoms. On the other hand in nonpolar covalent bonding, the electrons are equally shared and no charges are created.

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4 years ago

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][CO_3^{2-}]$

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

4 0

3 years ago

What type of charge does each part of an atom have (positive, negative, no charge)

Masteriza [31]

Answer:
P-positive
N-negative
E-no charge

6 0

3 years ago

Read 2 more answers

A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 31.μmol of copper(II) sulfate into a 150.mL volumetri

baherus [9]

Answer:

The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

Explanation:

The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.

Here's that idea written as a formula: c= n/V

where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.

You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L

Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L

Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM

3 0

2 years ago

12. Your client is prescribed Lipitor for high cholesterol, but has trouble swallowing a whole pill. How do you know Lipitor is

natima [27]

Answer:

The answer is C

Explanation:

This is the answer because it is the only way that it is okay to split in half.

3 0

3 years ago