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shutvik [7]
3 years ago
15

What kind of thermochemical equation is represented below?

Chemistry
1 answer:
ryzh [129]3 years ago
6 0
The  kind  of  thermochemical  equation  represented  below 

that  is
CaO(s)  + H2O (l)  =  Ca(OH)2 (s)  +65.2  kj

is exothermic  (  answer  B)
  This  is  because  its  heat  energy  has  a  +  sign  meaning  that  heat  is  released  by  the  reaction  above.
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From where in the solar system did scientists conduct their spectral anaylses in 1948? How do you know?
Kryger [21]

Answer:

They conducted it from earth

Explanation:

There had never been anything capable of observing it from space launched at that time because Sputnik wasn't even launched

3 0
3 years ago
Describe Rutherford's contribution to the atomic model.
Nady [450]

Answer:

Rutherford's experiment, also known as

\alpha  - scattering \: experiment

supports the existence of neutrons and the nucleus.

Explanation:

In the above diagram, Rutherford was trying to explain his contributions using thin foils of gold and other metals as targets for alpha particles from a radioactive source.

He observed that the majority of particles penetrated the foil either undeflected or with only a slight deflection. But, every now and then an alpha particle was scattered(or deflected) at a large angle..

According to Rutherford, most of the atoms must be empty space. This explains why the majority of alpha particles passed through through the gold foil with little or no deflection. The atoms positive charges, Rutherford proposed are all concentrated in the Nucleus, <em>which</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>dense</em><em> </em><em>central</em><em> </em><em>core</em><em> </em><em>withi</em><em>n</em><em> </em><em>the</em><em> </em><em>atom</em><em>. </em>

Whenever an alpha particle came close to a nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an alpha particle coming towards a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the Nucleus are called Protons.

I <em>hope</em><em> </em><em>you</em><em> </em><em>find</em><em> </em><em>this</em><em> </em><em>useful</em><em>.</em><em>.</em><em>. </em><em>Have</em><em> </em><em>a</em><em> </em><em>lovely</em><em> </em><em>day</em><em>. </em>

4 0
3 years ago
Given the balanced equation 2KC103+ 2KC1+302
8090 [49]

is it decomp single replacement double replacement

4 0
3 years ago
What is the final pressure (expressed in atm) of a 3.05 L system initially at 724 mm Hg and 298 K, that is compressed to a final
maksim [4K]

Hey there!:

V1 = 3.05 L

V2 = 3.00 L

P1 = 724 mmHg

P2 = to be calculated

T1 = 298 K

T2 = 273 K

Therefore:

P1*V1  / T1  = P2*V2 / T2

P2 = ( P1*V1 / T1  )   * T2 / V2

P2 = 724 * 3.05 * 273 / 298 * 3.00

P2 = 602838.6 / 894

P2 = 674.31 mmHg

1 atm ----------- 760 mmHg

atm ------------- 674.31 mHg

= 674.31 * 1 / 760

= 0.887 atm

Hope this helps!

5 0
3 years ago
A gas mixture is made up of kr (21.7 g), o2 (7.18 g), and co2 (14.8 g). the mixture has a volume of 23.1 l at 59 °c. calculate t
Scilla [17]

Answer:- partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm

Total pressure is 0.966 atm

Solution:- moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol

moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol

moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol

Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K

From ideal gas law equation, P = nRT/V

partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm

partial pressure of oxygen = (0.224 x 0.0821 x 332)/23.1 = 0.264 atm

partial pressure of carbon dioxide = (0.336 x 0.0821 x 332)/23.1 = 0.396 atm

Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm

4 0
3 years ago
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