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4 years ago

Which conclusion could be made from Ernest Rutherford’s gold foil experiment ?

1 answer:

mote1985 [20]4 years ago

3 0

From the Rutherford's gold foil experiment one can conclude that nucleus was very small in size, as compared to the atoms. In the experiment Rutherford discovered that, the atom contains a very small nucleus where all of its positive charge of the atom is present.

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Answer #4 and #5 make sure you have proof and will give brainliest

sweet-ann [11.9K]

Answer:

B,C

Explanation:

7 0

3 years ago

Read 2 more answers

From the list below, choose which groups are part of the periodic table

Nata [24]

The periodic table of the elements are describe the electronic configuration of the elements on which the properties of the elements depends. Among the given groups only metal, non-metal and semi-metal group are the part of periodic table. The metallic property depends upon the binding energy of the electrons with the nucleus. Thus the elements which have the valence electrons more near to the nucleus that is s-block elements are more metallic in nature. On the other hand the elements which have the valence electrons far from the nucleus are more non-metallic in nature like p-block elements. However the binding energy or the attraction of the outermost electrons to the nucleus depends not only its valence electrons position but also some other factors like shielding effect, effective nuclear charge etc.

The elements which are in between the metals and non-metals can be classified as semi-metals.

Although the conductivity of a material is an inherent property of the metals but sometime the nonmetals or semi-metals are also behave like a conductor due to presence of the other elements, thus it cannot be a p[property of the periodic table. Similarly acidity, flammable gases are not part of the periodic table.

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3 years ago

How many grams of potassium reacts with 16 grams of oxygen to produce 94 grams of potassium oxide

Anna71 [15]

Answer:

5.875

Explanation:

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3 years ago

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The wavelength of the violet light emitted from a hydrogen atom is 410.1 nm. This light is a result of electronic transitions be

VashaNatasha [74]

Answer:

e. 4.847 x 10-19 J

Explanation:

From the given information:

The equation connecting the photon energy and the wavelength is:

$E_{photon } = \dfrac {hc}{\lambda}$

where;

$planck's \ constant \ (h)$ = 6.626 * 10 ^{-34} J.s

$velocity \ of \ light \ (c)$ = 3.00 * 10^8 m/s

wavelength $\lambda = 410.1 \ nm \times \dfrac{10^{-9} \ m}{1 \ nm}$

$\lambda = 4.101 \times 10^{-7} \ m$

To determine the photon energy of violet light

$E_{photon } = \dfrac {(6.626 \times 10^{-34} J/s ) \times (3.00 \times 10^8 \ m/s) }{4.101 \times 10^{-7} \ m}$

= 4.847 × 10⁻¹⁹ J

4 0

3 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

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

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

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

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago