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The table lists the chemical formulas or symbols of a few common substances. Which two substances are compounds? A) aluminum and

katrin [286]

D is the correct answer

every other option contains an element

7 0

3 years ago

Read 2 more answers

calculate the number of moles of gas that occupy a 3.45L container at a pressure of 1.48 atm and a temperature of 45.6 Celsius

Otrada [13]

Answer:

There are 0,2 moles of gas that ocuppy the container.

Explanation:

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol. Firs we convert the unit of temperature in Celsius into Kelvin:

0°C= 273 K ------> 45,6 °C= 273 + 45, 6= 318, 6 K

PV= nRT ---> n= PV/RT

n= 1,48 atm x 3,45 L /0.082 l atm / K mol x 318,6 K

n= 0,195443479 mol

8 0

3 years ago

What is the new concentration of a solution of CaSO3 if 10.0 mL of a 2.0 M CaSO3 solution is diluted to 100 ml?

kifflom [539]

Answer: The new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

Explanation:

Given: $V_{1}$ = 10.0 mL, $M_{1}$ = 2.0 M

$V_{2}$ = 100 mL, $M_{2}$ = ?

Formula used to calculate the new concentration is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\10.0 mL \times 2.0 M = M_{2} \times 100 mL\\M_{2} = 0.2 M$

Thus, we can conclude that the new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

5 0

2 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

7 0

3 years ago

The C=C bond in 2-cyclohexenone produces an unusually strong signal. Explain using resonance structuresBased on the above resona

frosja888 [35]

Answer:

When two single single bonds separated by a double bond (e.g C=C-C=C or C=C-C=O in the case of 2-cyclohexenone), the effect of resonance among those there bonds will be observed.

Explanation:

Since the Oxygen atom has higher electronegativity, it will cause the electrons in the resonance bonds 'flow' toward the Oxygen atom, so that the C=C will 'lose' some electron. The signal read for that bond will be different from other alkene structure.

Attachment is the resonance structure of 2-cyclohexene.

8 0

3 years ago