An archaeologist discovers an ancient wooden bowl. which isotope would most likely be used to determine its age?

Commercially available hot packs are simple in design: a pouch with water on one side, isolated by a barrier from a specific sal

liq [111]

Answer:

b :)

Explanation:

an exothermic reaction is when heat/light is produced. heart is produced from this reaction so it is exothermic

3 0

1 year ago

(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: Enzyme is a protein-based molecule that speed up the rate of a reaction. Enzyme Kinetics studies the reaction rates of it.

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

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

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:

$0.25V_{max}=\frac{V_{max}[S]}{0.005+[S]}$

$0.25V_{max}(0.005+[S])=V_{max}[S]$

$0.00125+0.25[S]=[S]$

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:

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

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

3 0

1 year ago

At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)

viktelen [127]

Answer: $K_c$ for this reaction at this temperature is 0.029

Explanation:

Moles of $HBr$ = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of $HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M$

The given balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial conc. 0.500 M 0 M 0 M

At eqm. conc. (0.500-2x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}$

Equilibrium concentration of $[Br_2]$ = x = 0.0955 M

Now put all the given values in this expression, we get :

$K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}$

$K_c=0.029$

Thus $K_c$ for this reaction at this temperature is 0.029

7 0

2 years ago

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

Degger [83]

Answer : The % of (+) limonene isomer = 79%

The % of (-) limonene isomer = 0%

The % of enantiomeric excess = 58%

Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where, ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

${ee}=\frac{\%(+)-50\%}{2}$

$=\frac{79\%-50\%}{2}$

= 58%

7 0

2 years ago

Read 2 more answers

What is the formal charge on the nitrogen in hydroxylamine, h2noh?

andrew11 [14]

<h3>Answer:</h3>

Formal Charge on Nitrogen is "Zero".

<h3>Explanation:</h3>

Formal Charge on an atom in molecules is calculated using following formula;

Formal Charge = [# of Valence e⁻s] - [e⁻s in lone pairs + 1/2 # of Bonding e⁻s]

As shown in attached picture of Hydroxylamine, Nitrogen atom is containing two electrons in one lone pair of electrons and six electrons in three single bonds with two hydrogen and one oxygen atom respectively.

Hence,

Formal Charge = [5] - [2 + 6/2]

Formal Charge = [5] - [2 + 3]

Formal Charge = 5 - 5

Formal Charge = 0 (zero)

Hence, the formal charge on nitrogen atom in hydroxylamine is zero.

5 0

1 year ago