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ankoles [38]
3 years ago
12

a 6.7 volume of air, initially at 23 degrees celsius and .98 atm, is compressed to 2.7 L while heated to 125 degrees celsius. Wh

at is the final pressure?
Chemistry
1 answer:
Annette [7]3 years ago
6 0
Data:

V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm

V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?

Formula:

Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2

=> P2 = P1 V1 T2 / (T1 V2)

P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)

P2 = 3.22 atm
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8 0
3 years ago
Common brass is a copper and zinc alloy containing 37.0% zinc by mass and having a density of 8.48g/cm3. A fitting composed of c
inysia [295]
First, we calculate the mass of the sample:

mass = density x volume
mass = 8.48 x 112.5
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Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.

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3 0
3 years ago
Predict the missing product of this equation<br><br><br>1 MgF2 + 1 Li2CO3 -&gt; 1 ______ +2LiF
ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0
3 years ago
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