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Alona [7]
3 years ago
11

Write a summary paragraph discussing this experiment and the results. Use the following questions and topics to help guide the c

ontent of your paragraph.
Chemistry
1 answer:
erastovalidia [21]3 years ago
5 0

What is the experiment that is to be discuss/

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Which of the following particles is the smallest in radius
Gnesinka [82]

Answer:

Its helium

Explanation: Because teh effective nuclear charge holds the valence ectrons clos to the nuclues. the Atomic radius would decrease as you move from the led=ft and right of the perodic table

7 0
3 years ago
Read 2 more answers
What is the number of grams of N₂ (g) that contains 1.80 x 10^24 molecules?
svet-max [94.6K]

Answer:

83.69 gm

Explanation:

molar weight of N2 = 28

Find the number of moles then multiply by this

1.8 x10^24/ (6.022x10^23)   * 28 =83.69 gm

7 0
2 years ago
The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per uni
Nataliya [291]

Answer:

Explanation:

turn over number = R max / [E]t   = K2

From given , R max = 249 * 10 ^ -6 mol. L^-1

T [E]t = 2.23 n mol. L^-1

           =   2.23 * 10^-9 mol. L^-1

Putting values in above equation,

=          111.65 * 10^3           S^-1

Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.

5 0
3 years ago
An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 300 nm and then immediately emits photon whose ass
KIM [24]

Answer:

The net energy is 2.196 eV

Explanation:

Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.

Using:

ΔE = h*c*(1/λ_{1} - 1/λ_{2})

where:

ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*10^{-19} J.

h = 4.135*10^{-15} eVs

c = 3*10^{8} m/s

λ_{1} = 300 nm = 300*10^{-9} m

λ_{2} = 640 nm = 640*10^{-9} m

Thus:

ΔE = 4.135*10^{-15} eVs*3*10^{8} m/s*(\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m })

ΔE = 4.135*10^{-15}*3*10^{8}*1.77*10^{6} eV = 2.196 eV

6 0
3 years ago
The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc
Korvikt [17]

Answer:

The percentage deviation is  \Delta M = 1.87%

Explanation:

From the question we are told that  

     The concentration is of the solution is C = 1.0*10^{-5} M

     The true absorbance A = 0.7526

      The percentage of transmittance due to stray light z = 0.56% =\frac{0.56}{100}  = 0.0056

Generally Absorbance is mathematically represented as

           A = -log T

Where T is  the percentage of true transmittance

    Substituting value  

          0.7526 = - log T

              T = 10^{-0.7526}

                  = 0.177

                  = 17.7%

The Apparent absorbance is mathematically represented

           A_p = -log (T +z)

Substituting values

           A_p = -log(0.177 + 0.0056)

                = -log(0.1826)

               = 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

       \Delta A = \frac{A -A_p}{A} * \frac{100}{1}

              = \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}

             \Delta A = 1.87%  

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration  is

              \Delta M = 1.87%

           

6 0
3 years ago
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