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monitta
3 years ago
13

20 points will favor answer Please guys i could really use some help on this question

Chemistry
1 answer:
posledela3 years ago
8 0
Answer : Option B) <span>The oxygen in the air forms ozone in presence of UV rays

Explanation :
The other given options for the chemical change seems to be irrelevant as the option A which converts  water into different forms can be called as physical change, option C only dust and soot particle are released in air where no new substance is formed And in option D sulfur dioxide is getting released into air which is also a form of a physical process.

So in option B we can see a new compound getting formed which is ozone in presence of UV rays hence it is a chemical change.</span>
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What is the pH of a 0.028M solution (pH= -log(M))<br><br> A) 3.56<br> B) 2.88<br> C) 1.55<br> D) 1
kupik [55]

Answer:

1.55

Explanation:

-log(M)=pH

- Hope that helps! Please let me know if you need further explanation.

4 0
3 years ago
The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction
shepuryov [24]

Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

       NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)

(a)  When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of H_{2}S will increase.

  • (b)  When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of H_{2}S will decrease with decrease in volume.

  • When we increase the mount of NH_{4}HS then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

Thus, we can conclude that formation of H_{2}S will increase then.

3 0
3 years ago
What Element am I?
rjkz [21]

Answer:

The correct answer is - sulfur.

Explanation:

In the periodic table, there are 18 groups and 7 rows or periods arranged according to their atomic number or electronic configuration. In the question, it is mentioned that the desired element atomic mass is less than the atomic mass of the selenium which is 78.96, and more than oxygen which is 15.99 with 6 electron valence and present in the third row.

As it has 6 valency of electron it must be in the 16 group of the table that comprises the 6 valency and as it is located in the 3rd row it must be sulfur that also has an atomic mass between selenium and oxygen.

6 0
3 years ago
A cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the system is raise
Black_prince [1.1K]

Answer:

There is 96200 J or 96 kJ of heat released.

Explanation:

<u>Step 1: </u>Data given

Mass of helium = 250 grams

Temperature = 200 K

Temperature is raised by 74 K

The heat capacity of helium = 20.8 J/mol*K

Molecular weight for helium = 4 g/mol

<u>Step 2:</u> Calculate moles of Helium

Moles of Helium = mass of Helium / molar mass of helium

Moles of helium = 250 grams / 4g/mol

Moles of helium = 62.5 moles

<u>Step 3:</u> Calculate heat

Q = n*c*ΔT  or

⇒ with Q = the heat released in Joule

⇒ with n = moles of helium = 62.5 moles

⇒ with c = the heat capacity of helium = 20.8 J/mol*K

⇒ with ΔT = the change in temperature = T2 - T1 = 74 K

Q = 62.5 mol * 20.8 J/mol*K * 74 K

Q = 96200 J = 96 kJ

Since the temperature is raised, this is an exothermic reaction. The heat is released.

There is 96200 J or 96 kJ of heat released.

8 0
3 years ago
Please help me I need this now I'm from Phillippines it's science subject<br>​
Lorico [155]
I know the first answer is homogenous
7 0
3 years ago
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