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shusha [124]
3 years ago
13

The gas in a container is at a pressure of 3.00 atm at 25.0 C. Directions on the container warn the user no to keep it in a plac

e where the temperature exceeds 52 C. What would be the gas pressure in the container at 52 C?
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

P_{2} = 3.272\,atm

Explanation:

Let consider that gas behaves ideally, the following relation is used:

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}

The final pressure is:

P_{2} = P_{1} \cdot \frac{T_{2}}{T_{1}}

P_{2} = (3\,atm)\cdot \left(\frac{325.15\,K}{298.15\,K}  \right)

P_{2} = 3.272\,atm

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Why do some elements form double and triple bonds during bonding
jekas [21]

Explanation:

Elements need a total of eight electrons to gain stability and look like a noble gas. So, they sometimes need sharing of two, four or even six electrons to complete their octate. So, they form double and triple covalent bonds. One more  the reason is the  interaction between the p orbitals of the combining atoms. for example  A double bond, as in ethene H2C=CH2, arises from one combination of the s orbitals and one combination of the p_y orbitals.

4 0
3 years ago
How many atoms are in 0.650 mole of zinc?<br> NEED TO KNOW ASAP PLEASE
ryzh [129]
There are
4.517
⋅
10
23
atoms of Zn in 0.750 mols of Zn.
5 0
3 years ago
How many moles of acetylene are needed to completely react with 16.5g of CO2
kykrilka [37]

Answer:

0.375 mols

Explanation:

  1. Relative atomic mass Ar of C=12×1=12
  2. Relative atomic mass Ar of O=16×2=32
  3. Realitive formula mass Mr is 12+32=44

mol:mass

1 : 44

x :16.5

cross multiply and get the answer

5 0
3 years ago
your friend offers to show you an intrusive igneous rock. which of the following would you expect to see?
Scorpion4ik [409]
You would expect to see a marble-looking rock
3 0
4 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
3 years ago
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