Do only chemical reactions get colder/hotter?

Which model of the atom was used as a result of JJ Thomson's cathode ray

mariarad [96]

Answer:

a. Plum pudding model

Explanation:

The plum pudding model of the atom was proposed by J.J. Thomson. It was the model he derived from his experiment on the gas discharge tube.

J.J Thomson was the first person to discover electrons which he called cathode rays because in the discharge tube, they emanate from the cathode.

This led him to suggest the plum pudding model of the atom.
The model reflects electrons being surrounded by a volume of negative charges.

6 0

3 years ago

Which of the following is supported by reliable evidence?

Tasya [4]

B. Theory of gravity

3 0

2 years ago

Do all substances absorb heat the same way?

miv72 [106K]

Since particles are closer together, solids conduct heat better than liquids or gases. Conduction moves heat through a material. It keeps a fire going by spreading the heat through solid material. Radiation is a method of heat transfer that does not require particles to carry the heat energy.

5 0

3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

Allisa [31]

Answer:

The mass of water $m_{w}$ = 39.18 gm

Explanation:

Mass of iron $m_{iron}$ = 32.5 gm

Initial temperature of iron $T_{1}$ = 22.4°c = 295.4 K

Specific heat of iron $C_{iron}$ = 0.448 $\frac{KJ}{kg K}$

Mass of water = $m_{w}$

Specific heat of water $C_{w} = 4.2 \frac{KJ}{kg K}$

Initial temperature of water $T_{2}$ = 336 K

Final temperature after equilibrium $T_{f}$ = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water = Heat gain by iron rod

$m_{w}$ $C_{w}$ ( $T_{2}$ - $T_{f}$ ) = $m_{iron}$ $C_{iron}$ ( $T_{f}$ - $T_{1}$ )

Put all the values in above formula we get

$m_{w}$ × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

$m_{w}$ = 39.18 gm

Therefore the mass of water $m_{w}$ = 39.18 gm

8 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago