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riadik2000 [5.3K]
3 years ago
6

What are the sub-particles that make up an atom? (select all that apply) *?

Chemistry
1 answer:
Mrac [35]3 years ago
7 0
The answer should be neutrons electrons and protons
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Phosphorus trichloride (PC) is used as a reactant in the production of many other chemicals that contain phosphorus, including p
goldenfox [79]

Answer:

137.33 g/mol

Explanation:

6 0
2 years ago
[40 Points]
pashok25 [27]

the one that is wrong is a. increase the molecular weight of the gas , the molecular weight never changes when a solid turns to a liquid or when a liquid turns to a gas .

8 0
3 years ago
Determine the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0 C containing a mixture of xenon and n
lawyer [7]

Answer:

The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981

Explanation:

Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761

As we know,

\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239

According to Dalton's Law of partial pressure-

P_i=\chi_i\times P_{total}

Where,

P_i=The pressure of the gas component in the mixture

\chi_i= Mole fraction of that gas component

P_t= The total pressure of the mixture

P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm

<u>Calculation: </u>

To calculate the number of moles,

PV=nRT

n=\frac{PV}{RT}

n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole

Learn more about Dalton's Law of partial pressure here;

brainly.com/question/14119417

#SPJ4

4 0
1 year ago
During photosynthesis, plants transform
attashe74 [19]
During Photosynthesis , Plants Transforms Light Energy Into Chemical Energy .
6 0
3 years ago
What is the value for the reaction: N2(g) + 2 O2(g) --&gt; N2O4(g) in terms of K values from the reactions:
Valentin [98]

Answer : The correct expression will be:

K=(K_1)^2\times K_2

Explanation :

The chemical reactions are :

(1) \frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons NO(g) K_1

(2) 2NO(g)+O_2(g)\rightleftharpoons N_2O_4(g) K_2

The final chemical reaction is :

N_2(g)+2O_2(g)\rightleftharpoons N_2O_4 K=?

Now we have to calculate the value of K for the final reaction.

Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.

If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.

If the two equations are added then equilibrium constant will be multiplied.

Thus, the value of 'K' will be:

K=(K_1)^2\times K_2

6 0
3 years ago
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