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3 years ago

What did Ernest Rutherford s gold foil experiment demonstrate about atoms?

1 answer:

mote1985 [20]3 years ago

6 0

Rutherford performed gold foil experiment to understand that how negative and positive particles could Co exist in an atom. He bombarded alpha particles on a 0.00004 cm thick gold foil.

He proposed a planetary model of the atom and concluded following results and demonstrated that,
1. An atom produces a line spectrum.
2. An Electron revolves around the nucleus without any orbits.
3. Since most of the particles passed through the foil undeflected it means that most of the volume occupied by an atom is empty.
4. An Atom as a whole is neutral.
5. The deflection of few particles on the foil suggested that there is center of positive particles in an atom called the nucleus of the atom.
6. The complete rebounce of few particles on the gold foil suggested that the nucleus is very dense and hard.

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There are two types of nucleic acids, dna and rna. nearly all organisms use dna, not rna, as the central repository for genetic

Leya [2.2K]

Found the choices. Pls see attachment.

The statements that explains this phenomenon are:
1) DNA contains adenine as one of its nitrogenous bases.
2) DNA has a double-stranded structure that ensures an accurate mechanism of duplication.

6 0

2 years ago

Convert 15.6 g Na2Cr2O7 to moles of Na2Cr2O7.

Leviafan [203]

Answer:

261.96754

Explanation:

3 0

3 years ago

0.350 mol of a solid was dissolved in 260 mL of water at 21.2 oC. After the solid had fully dissolved, the final temperature of

Fittoniya [83]

Answer: Heat of the solution = mass water × specific heat water × change in temperature

mass water = 260ml (1.00g/ml ) = 260g

specific heat of water = c(water) = 4.184J/ g°C

Heat change of water = final temperature - initial temperature

= 26.5 - 21.2

= 5.3 °C

H = 260 g ( 4.184J/g°C ) (5.3°C) = 5765J

Molar heat = $\frac{5765J}{0.350mol}$

= 16473J/mol

Explanation: finding molar heat requires first to look at specific heat of water and the change of water temperature

7 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

Help me please, will give brainliest if correct!

mafiozo [28]

The answer would be C

4 0

3 years ago

Read 2 more answers