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IRISSAK [1]
1 year ago
14

zn (s) 2 hcl (aq) -----> zncl2 (aq) h2(g) if 520 ml of h2 is collected over water at 28oc and the atmospheric pressure is 1.0

atm, how many g of zn was used at the start of the reaction? (vapor pressure of water at 28oc is 28.3 mmhg).
Chemistry
1 answer:
nadya68 [22]1 year ago
3 0

The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.

Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.

We recognise that

h * 2 = PT - P * h * 20 = 1atm - 0.037atm

= 0.963 atm

1 * h * 2 = Ph * 2V / R * T

= 0.963 atm x 0.520 L / 0.0821 L atm/

molK * 301

= 0.02 mol h2

= 0.02molZn

So 0.02 mol Zn x 65.39 g/mol

= 1.33 g Zn

Read more about zinc;

brainly.com/question/28880469

#SPJ4

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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
3 years ago
How strongly gravity pulls on an objects mass is the object
Lemur [1.5K]
Yes ...................................
5 0
3 years ago
A large cyclotron directs a beam of He++ nuclei onto a target with a beam current of 0.250 mA. (a) How many He++ nuclei per seco
nikitadnepr [17]

Answer:

a. 7.8*10¹⁴ He⁺⁺ nuclei/s

b. 4000s

c. 7.7*10⁸s

Explanation:

I = 0.250mA = 2.5 * 10⁻³A

Q = 1.0C

1 e- contains 1.60 * 10⁻¹⁹C

But He⁺⁺ Carrie's 2 charge = 2 * 1.60*10⁻¹⁹C = 3.20*10⁻¹⁹C

(A).

No. Of charge per second = current passing through / charge

1 He⁺⁺ = 2.50 * 10⁻⁴ / 3.2*10⁻¹⁹C

1 He⁺⁺ = 7.8 * 10¹⁴ He⁺⁺ nuclei

(B).

I = Q / t

From this equation, we can determine the time it takes to transfer 1.0C

I = 1.0 / 2.5*10⁻⁴ = 4000s

(C).

Time it takes for 1 mol of He⁺⁺ to strike the target =?

Using Avogadro's ratio,

1.0 mole of He = (6.02 * 10²³ ions/mol ) * (1 / 7.81*10¹⁴ He ions)

Note : ions cancel out leaving the value of the answer in mols.

1.0 mol of He = 7.7 * 10⁸s

8 0
3 years ago
The table below gives the masses of two different subatomic particles found in an atom.
kap26 [50]
I think the subatomic particles that are paired with each of their corresponding name would be :
1. X , proton and z, electron

hope this helps
7 0
3 years ago
Read 2 more answers
Displacement is the slope of a velocity vs. time graph<br><br> A. True<br><br> B. False
mezya [45]

Answer:

The given statement is false.

Explanation:

Displacement is a vector quantity which represent shortest distance form initial position to final position.

Velocity of an object is defined as ratio of displacement of object to that of the time taken by object to move from its initial position to final position.

In displacement vs time graph the slope of the graph represent the velocity. Whereas slope of velocity vs time graph gives acceleration.

6 0
3 years ago
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