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ss7ja [257]
3 years ago
6

How many molecules of nitrogen dioxide are in 1.28 g of nitrogen dioxide?

Chemistry
2 answers:
Kryger [21]3 years ago
5 0
In order to determine the number of molecules of nitrogen dioxide that are in 1.28 g of nitrogen dioxide, we first need to know these values:

Molar mass of nitrogen dioxide = 46.0055 g/mol
Avogadro's number = 6.022 x 10^23 molecules/mol

Then, we divide 1.28 g by the molar mass and multiply the resulting value to Avogadro's number:

1.28 g/ 46.0055 g/mol x 6.022 x 10^23 molecules/mol = 1.6755 x 10^22 molecules of nitrogen dioxide
puteri [66]3 years ago
3 0

Hello! 

How many molecules of nitrogen dioxide are in 1.28 g of nitrogen dioxide?

Let's first find the molecular mass of nitrogen dioxide, knowing that by Avogadro's Law for each mole of a substance we have 6.02 * 10²³ molecules.

N = 1*14 = 14 amu

O = 2*16 = 32 amu

---------------------------

molecular mass of nitrogen dioxide = 14 + 32 = 46 g/mol

How many molecules of nitrogen dioxide are in 1.28 g of nitrogen dioxide NO2?

46 g ---------------- 6.02*10²³ molecules

1.28 g -------------  y molecules

46*y = 1.28*6.02*10^{23}

46y = 7.7056*10^{23}

y = \dfrac{7.7056}{46}*10^{23}

y = 0.167513043*10^{23}

\boxed{\boxed{y \approx 1.675*10^{22}\:molecules\:of\:nitrogen\:dioxide}}\end{array}}\qquad\checkmark

________________________

I Hope this helps, greetings ... Dexteright02! =)

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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
2 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
3 years ago
You collect 552 mL of argon gas at 23.0 C. What volume will the gas occupy at 46.0C if the pressure remains constant?
taurus [48]

Answer:

1027.9 mL

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

8 0
2 years ago
PLEASE HELP it would be really nice of you to answer this question i would be very grateful &lt;3
Aleksandr-060686 [28]

Answer: D (The abundance percentage of each isotope)

Explanation: hope this helps!

7 0
3 years ago
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Hunter-Best [27]

Answer:

1) correct

2) incorrect

3) correct

4)incorrect

Explanation:

1) A Lewis acid is a substance that accepts a nonbonding pair of electrons.

A Bronsted-Lowry acid is a substance that donates a proton H⁺

Since the donation of a proton involves the acceptance of a pair of electrons, every Bronsted-Lowry acid is also a Lewis acid.

2)A Lewis acid not necessarily needs to have a proton to be donated.

3) Conjugated acids of weak bases are strong acids and conjugated acids of strong bases are weak acids.

4)K⁺ comes from a strong base, therefore is does not have an acidic behaviour.

4 0
2 years ago
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