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egoroff_w [7]
3 years ago
7

Based on the volumes of solutions you mixed together for your six kinetics runs, you have found the three runs where the [H3O+ ]

is constant. The values for [Sn2+ ], and kexpt are as follows: Species: Run a , Run b , Run c [Sn2+ ]: 0.0063 , 0.0083 , 0.0123 kexpt : 2.96e−7 , 6.69e−7 , 2.14e−6 Plot in Excel on the x-axis the ln [Sn2+ ] and on the y-axis the ln kexpt and fit it to a linear trend line in Excel. Input in the field below what is exponent that [Sn2+ ] is raised to in the rate law expression (this is the slope of this fit - write in exactly what Excel tells you the slope is).
Chemistry
1 answer:
dybincka [34]3 years ago
6 0
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Calculate the number of water (H2O) molecules produced from the decomposition of 75.50 grams of Iron (III) hydroxide (Fe(OH)3).
Phantasy [73]

Answer: 6.38\times 10^{23} molecules of water are produced.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe(OH)_3=\frac{75.50g}{106.8g/mol}=0.707moles

The balanced chemical reaction is:

2Fe(OH)_3\rightarrow Fe_2O_3+3H_2O  

According to stoichiometry :

2 moles of Fe_2O_3 produce = 3 moles of H_2O

Thus 0.707 moles of Fe_2O_3 will produce=\frac{3}{2}\times 0.707=1.06moles  of H_2O  

According to avogadro's law, 1 mole of every substance contains avogadro's number 6.023\times 10^{23} of particles.

Thus 1.06 moles of H_2O contains = \frac{6.023\times 10^{23}}{1}\times 1.06=6.38\times 10^{23} molecules

5 0
3 years ago
The electronegativity of the element affects all of the following except a- type of bond it forms with other elements b- the boi
seraphim [82]

The electronegativity of the element affects the ability of its compounds to dissolve in different solvents.

<h3>What is the meaning of electronegativity?</h3>

Electronegativity is a measure of an atom's ability to attract shared electrons to itself.

Polar bonds have a positive and negative side to them, and therefore can attract water dipoles and dissolve in water.

If the polarities of the solvent and solute match (both are polar or both are nonpolar), then the solute will probably dissolve.

If the polarities of the solvent and solute are different (one is polar, one is nonpolar), the solute probably won't dissolve.

Hence, option D is correct.

Learn more about  electronegativity here:

brainly.com/question/14560699

#SPJ1

6 0
2 years ago
Which type of matter can be separated by physical means? A. atom B. element C. compound D. mixture
madreJ [45]

Answer:

b

Explanation:

A pure substance has a constant composition and cannot be separated into simpler substances by physical means. There are two types of pure substances: elements and compounds.

7 0
2 years ago
Read 2 more answers
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 specific silver is How many joules of energy are needed to warm 4.37 g of silver from 25.0 degrees * C to 27.5 degrees * C ?
Alexxandr [17]

0.24J/g*degC * 4.37g * 2.5degC = 2.622J

The 2.5 degC is the difference between 25 and 27.5 deg C.

6 0
3 years ago
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