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Ivanshal [37]
3 years ago
13

What would happen to the entropy in the reaction 2O^3(9) ► 3O^2 (g)?

Chemistry
1 answer:
avanturin [10]3 years ago
7 0
It is B I think but I’m not completely sure
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If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?
Llana [10]
Since the temperature is constant, therefore, this problem can be solved based on Boyle's law.
Boyle's law states that: " At constant temperature, the pressure of a certain mass of gas is inversely proportional to its pressure".

This can be written as:
P1V1 = P2V2
where:
P1 is the initial pressure = 1 atm
V1 is the initial volume = 3.6 liters
P2 is the final pressure = 2.5 atm
V2 is the final volume that we need to calculate

Substitute with the givens in the above mentioned equation to get the final volume as follows:
P2V1 = P2V2
1(3.6) = 2.5V2
3.6 = 2.5V2
V2 = 3.6 / 2.5 = 1.44 liters
8 0
3 years ago
Read 2 more answers
SOMEONE HELP ME PLEASE ILL GIVE BRAINLY
Maksim231197 [3]

Answer:

true

Explanation:

just did it

6 0
3 years ago
A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the m
aliya0001 [1]

Answer:

311.25k

Explanation:

The question assumes heat is not lost to the surroundings, therefore

heat emitted from hotter sample ( q_{\ lost} )= heat absorbed by the less hotter  sample( q_{\ gain} )

The relationship between heat (q), mass (m) and temperature (t) is q = mc\Delta t

where c is specific heat capacity, \Delta t temperature change.

\Delta t = t_{\ final} - t_{\ initial}

equating both heat emitted and absorb

-q_{\ lost} = q_{\ gain}

-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})

where the values with subset 1 are the values of the hotter sample of water and the values with subset 2 are the values of the less hot sample of water.

C will cancel out since both are water and they have the same specific heat capacity.

so we have

-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})

where m1 = 50g, t 1initial = 330, m2 = 30g, t2 initial = 280,t final (final temperature of the mixture) = ?

-50 * (t_{final} - 330) = 30 *  (t_{final} - 280)

-50t_{final} + 16500 = 30t_{final} - 8400

80t_{final} = 16500+8400

80t_{final} = 24900

t_{final} = 24900/80 = 311.25k

8 0
3 years ago
25 POINTS TO WHOEVER CAN FILL THIS OUT! Radioactivity Decay
Degger [83]

Answer:

how long?

Explanation:

3 0
4 years ago
In the reaction of aluminum and iron(iii) oxide to form iron and aluminum oxide, δh is - 850 kj. how many grams of aluminum oxid
Semmy [17]
The  grams  of   aluminium  oxide  are  formed when  350  kj  heat  are  released is calculated  as  follows

1mole  =  850Kj
what  about  350kj
=1mole  x350/850  =  0.412  moles

mass  of  Al =   moles  of  Al x  molar  mass of  Al

=  0.412mol  x  27  g/mol  =  11.124  grams
8 0
3 years ago
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