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2 years ago

Annalise observes a streak of light in the sky and says she is seeing a shooting star. What is she most likely observing?

Chemistry

2 answers:

Katena32 [7]2 years ago

7 0

Answer: A Meteor

Explanation:

Phoenix [80]2 years ago

5 0

Answes:

c) a meteor

Explanation:

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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 = $\frac{11V}{10}$

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

A solid sample of a compound and a liquid sample of the same compound are each tested for electrical conductivity. which test co

marshall27 [118]

If these are you choices:
(1) Both the solid and the liquid are good conductors.

(2) Both the solid and the liquid are poor conductors.

(3) The solid is a good conductor, and the liquid is a poor conductor.

(4) The solid is a poor conductor, and the liquid is a good conductor.

Then the answer is number 4. This is because ionic compound conducts electricity when it is dissolved in water.

8 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

2 years ago

A) A cat dish has a mass of 1975 g and a volume of 7500ml. What is the overall density of the cat dish? B) find the maximum mass

maxonik [38]

What's the answer? It asked to be 20 characters long so just writing this.

5 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:

Moles Mg²⁺:

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

Moles B(C₆H₄)₄ = Moles Lithium:

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 =

4 0

2 years ago