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1 year ago

5

Rutherford, geiger, and marsden’s experiment demonstrated that the volume of the nucleus is roughly what fraction of the volume

occupied by the electrons?

1 answer:

trasher [3.6K]1 year ago

4 0

Rutherford, Geiger and Marsden's experiment proved that every atom has a nucleus and that this nucleus is of positive charge and contains the most of the mass of the atom. 0.005% of the volume occupied by the electrons is the volume of the nucleus.

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A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

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1 year ago

In this lab, you will do experiments to identify types of changes. Using the question format you learned (shown above), write an

Vlad1618 [11]

Answer:

How can you distinguish a physical change from a chemical change?

Explanation:

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1 year ago

Salt solutions can be __________ to give solid salts. What word completes this sentence?

NikAS [45]

Answer:

evaporated

Explanation:

Once the solution is evaporated there will only be salt left, since the only other thing in the soulution is water.

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2 years ago

During a process called photoact, ________ give up an electron as a part of the light-dependent reactions.

dybincka [34]

Answer:

Chloroplasts?

Explanation:

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1 year ago

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w

max2010maxim [7]

Explanation:

The given data is as follows.

$\lambda$ = 253.4 nm = $253.4 \times 10^{-9}m$ (as 1 nm = $10^{-9}$ )

$n_{1}$ = 5, $n_{2}$ = ?

Relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}$

= $0.0784 \times 10^{-17}$ J

= $7.84 \times 10^{-19} J$

Hence, energy released is $7.84 \times 10^{-19} J$ .

Also, we known that change in energy will be as follows.

$\Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}$

where, Z = atomic number of the given element

$7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

$\frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

0.02 + 0.04 = $\frac{1}{n^{2}_{1}}$

$n_{1}$ = $\sqrt{\frac{1}{0.06}}$

= 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0

1 year ago