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3 years ago

What led Thomson to decide that Dalton's atomic model needed to be revised?

Chemistry

1 answer:

jeka943 years ago

4 0

Answer:

Discovery of electron while studying the properties of cathode ray by Thomson suggested that Dalton atomic model should be revised.

Explanation:

Electron was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

He constructed the glass tube and create vacuum in it. He applied electric current between electrodes. He noticed that a ray of particles coming from cathode to wards positively charged anode. This ray was cathode ray.

Properties of cathode ray:

The ray is travel in straight line.

The cathode ray is independent of composition of cathode.

When electric field is applied cathode ray is deflected towards the positively charged plate.

Hence it was consist of negatively charged particles.

Symbol= e-

Mass= 9.10938356×10-31 Kg

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

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Balloon A started with a volume of 1.76 L. The temperature of the room the balloon was in was 295 K. The balloon was heated to a

valentina_108 [34]

Answer:

V₂ = 1.5 L

Explanation:

Given data:

Initial volume of balloon = 1.76 L

Initial temperature = 295 K

Final temperature = 253.15 K

Final volume = ?

Solution:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1.76 L ×253.15 K / 295 K

V₂ = 445.54 L.K /295 K

V₂ = 1.5 L

4 0

3 years ago

In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o

Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

$V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1} )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w$

The specific gravity of the wood chips is 0.494.

The average volume of a log is

$V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2} \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}$

The weight of one log is

$M=\rho*V=0.494*\rho_w*12.57 ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm$

To provide 3617 ton/day of wood chips, we need

$n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min$

The feed rate of logs is 14.3 logs/min.

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