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3 years ago

From his experiments, J. J. Thomson concluded that

Chemistry

2 answers:

In-s [12.5K]3 years ago

7 0

atoms contain small negatively charged particles that are called electrons.

sergiy2304 [10]3 years ago

6 0

Answer: option D. Atoms containd small negatively charged particles that are called electrons.

JJ Thompson carried out three important experiments with cathode rays. With them he proveed that cathode rays were negative charged particles and that the ratio charge/mass was so large that the particles either had a huge charge or had a very small mass, and he came up with the latter, defining that these negative charged particle were very small.

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A student has a solution with a pH of 7. The student adds HF, hydrogen floride, to the solution, hoping to increases the pH of t

Allushta [10]

Low pH = high acidity.
HF has a very low pH, so when added to a solution, it will lower the pH of the solution and therefore make it more acidic. So the answer should be B.

5 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [

Setler [38]

Answer:

$rate=-1.75x10^{-5}\frac{M}{s}$

Explanation:

Hello,

In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

$rate=\frac{1}{2}*\frac{C_f-C_0}{t_f-t_0}$

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript f is referred to the final condition and the subscript 0 to the initial condition, thus, we obtain:

$rate=\frac{1}{2}*\frac{0.00650M-0.0100M}{100s-0s}\\\\rate=-1.75x10^{-5}\frac{M}{s}$

Clearly, it turns out negative since the concentration is diminishing due to its consumption.

Regards.

3 0

3 years ago

A ____________ is a large body that moves around a star?

Travka [436]

Answer:

A Planet

Explanation:

The earth for example, is a large body that orbits the sun, our local star

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3 years ago

Read 2 more answers

What law is known as the law of action-reaction

Marianna [84]

Answer:

It would Newton's third law.

Explanation:

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3 years ago