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Yuliya22 [10]
2 years ago
10

Plz help this is due today

Chemistry
2 answers:
vazorg [7]2 years ago
7 0

Answer:

Explanation:

-

DENIUS [597]2 years ago
5 0

Hello there, I hope you are having a great day Your Question: Which is not an advantage of sexual reproduction? Would be A Diverse Offspring because some could be got to the mother or father carry are mixed together in the offspring.

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You divide 19 centimeters by 2.54 making 19 centimeters equivalent to 7.48 inches
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White gold is used in jewelry and contains two elements, gold and palladium. A jeweler has two different samples that are both i
wlad13 [49]
Solution : As they have uniform composition throughout they are considered as homogeneous mixture. Both samples are mixture of two metals (gold and palladium) thus are alloys.
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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined
Sunny_sXe [5.5K]

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here,  suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.  

3 0
3 years ago
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3 years ago
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