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3 years ago

Role and significance of guassion or normal distribution?

Chemistry

2 answers:

Lesechka [4]3 years ago

7 0

Answer:

The normal distribution (or Gaussian distribution), also referred as bell curve, is very useful due to the central limit theorem. Normal distribution states which are average of random variables converge in distribution to the normal and are normally distributed when the number of random variables is large.

kkurt [141]3 years ago

5 0

Normal my guy!!!!!!!!!

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An aqueous solution contains 0.100 m naoh at 25.0 °c. the ph of the solution is ________.

ikadub [295]

Recall; pH + pOH = 14
In this case [OH-] =0.100 m
therefore;
pOH = -LOG[OH-]
= - Log (0.100)
= 1.00
Therefore; the pOH is 1.00
And since, pH +pOH = 14
Then pH = 14-pOH
= 14 -1
= 13
Thus the pH is 13.00

7 0

3 years ago

One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu

Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0

2 years ago

At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubil

8090 [49]

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

$\frac{S_{1}}{P_{1}}=\frac{S_{2}}{P_{2}}$

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

$\frac{0.725}{3.08}=\frac{S_{2}}{8}$

On rearranging,

$S_{2}=\frac{0.725\times 8}{3.08}=1.88$

Therefore, solubility will be 1.88 mg of $N_{2}$ gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.

3 0

2 years ago

Which isotope will spontaneously decay and emit particles with a charge of +2?

VikaD [51]

Answer: The correct option is 4.

Explanation: All the options will undergo some type of radioactive decay processes. There are 3 decay processes:

1) Alpha decay: It is a decay process in which alpha particle is released which has has a mass number of 4 and a charge of +2.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

2) Beta-minus decay: It is a decay in which a beta particle is released. The beta particle released has a mass number of 0 and a charge of (-1).

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

3) Beta-plus decay: It is a decay process in which a positron is released. The positron released has a mass number of 0 and has a charge of +1.

$_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0\beta$

For the given options:

Option 1: This nuclei will undergo beta-plus decay process to form $_{25}^{53}\textrm{Mn}$

$_{26}^{53}\textrm{Fe}\rightarrow _{25}^{53}\textrm{Mn}+_{+1}^0\beta$

Option 2: This nuclei will undergo beta-minus decay process to form $_{80}^{198}\textrm{Hg}$

$_{79}^{198}\textrm{Au}\rightarrow _{80}^{198}\textrm{Hg}+_{-1}^0\beta$

Option 3: This nuclei will undergo a beta minus decay process to form $_{56}^{137}\textrm{Ba}$

$_{55}^{137}\textrm{Cs}\rightarrow _{56}^{137}\textrm{Ba}+_{-1}^0\beta$

Option 4: This nuclei will undergo an alpha decay process to form $_{85}^{216}\textrm{At}$

$_{87}^{220}\textrm{Fr}\rightarrow _{85}^{216}\textrm{At}+_2^4\alpha$

Hence, the correct option is 4.

7 0

3 years ago

Read 2 more answers

Periodic table, chemistry questions

ehidna [41]

Answer:

Explanation:

The Lanthanides were first discovered in 1787 when a unusual black mineral was found in Ytterby, Sweden. This mineral, now known as Gadolinite, was later separated into the various Lanthanide elements. In 1794, Professor Gadolin obtained yttria, an impure form of yttrium oxide, from the mineral.

7 0

3 years ago