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Schach [20]
3 years ago
7

What is a homogeneous mixture?

Chemistry
2 answers:
Alina [70]3 years ago
8 0

Answer:

A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture

Explanation:

Examples:

  • Air
  • Coffee
  • Steel
  • Blood
  • Vinegar
Marrrta [24]3 years ago
3 0

Answer:

Explanation:

A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture. The salt water described above is homogeneous because the dissolved salt is evenly distributed throughout the entire salt water sample.

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Two reasons why emerging diseases are especially harmful to humans. (is science)
Aleonysh [2.5K]

Answer:

Many emerging diseases arise when infectious agents in animals are passed to humans (referred to as zoonoses). As the human population expands in number and into new geographical regions, the possibility that humans will come into close contact with animal species that are potential hosts of an infectious agent increases.

Explanation:

7 0
2 years ago
What will the pressure be if 89.9 moles of argon are contained in a 12.0 L cylinder that is pressurized at a temperature of 300
irinina [24]
  • P=nRT/V
  • p=89.9(8.314)(12)/300
  • P=8969.14/300
  • P=29.89atm
  • P=29.9atm

Done!

7 0
2 years ago
What is the mass of a 2.5 mole sample of MgO2
MrRa [10]
Mass = no. of moles x molecular weight
m = n x Mr
m = 2.5 mol x (24 + [16 x 2])
m = 140g
6 0
2 years ago
What about a sponge? it is a solid yet we are able to compress it. why?
erastova [34]

Answer:

ponge is a solid yet we able to compress it because it has holes in it. When we compress the sponge the air goes out of those pores.

Explanation:

6 0
3 years ago
Read 2 more answers
One mole of an ideal gas, CP=3.5R, is compressed adiabatically in a piston/cylinder device from 2 bar and 25 oC to 7 bar. The pr
Luda [366]

Answer:

the entropy change of the gas is ΔS= 2.913 J/K

Explanation:

starting from the first law o thermodynamics for an adiabatic reversible process

ΔU= Q - W

where

ΔU = change in internal energy

Q= heat flow = 0 ( adiabatic)

W = work done by the gas

then

-W=  ΔU

also we know that the ideal compression work Wcom= - W , then Wcom = ΔU. But also for an ideal gas

ΔU= n*cv* (T final - T initial)

where

n=moles of gas

cv= specific heat capacity at constant volume

T final =T₂= final temperature of the gas

T initial =T₁= initial temperature of the gas

and also from an ideal gas

cp- cv = R → cv = 7/2*R - R = 5/2*R

therefore

W com = ΔU = n*cv* (T final - T initial)

for an ideal gas under a reversible adiabatic process ΔS=0 and

ΔS= cp*ln(T₂/T₁) - R* ln (P₂/P₁) =0

therefore

T₂ = T₁* (P₂/P₁)^(R/cp) = T₁* (P₂/P₁)^(R/(7/2R))=  T₁* (P₂/P₁)^(2/7)

replacing values T₁=25°C= 298 K

T₂ =T₁* (P₂/P₁)^(2/7)  = 298 K *(7 bar/2 bar)^(2/7) = 426.25 K

then

W com = ΔU = n*cv* (T₂- T₁)  

and the real compression work is W real = 1.35*Wcom , then

W real = ΔU

W real = 1.35*Wcom = n*cv* (T₃ - T₁)

T₃ = 1.35*Wcom/n*cv + T₁ = 1.35*(T₂- T₁) + T₁ =1.35*T₂ - 0.35*T₁ = 1.35*426.25 K - 0.35 *298 K = 471.14 K

T₃ = 471.14 K

where

T real = T₃  

then the entropy change will be

ΔS= cp*ln(T₃/T₁) - R* ln (P₂/P₁) = 7/2* 8.314 J/mol K *ln(471.14 K /298 K ) - 8.314 J/mol K* ln (7 bar / 2 bar)  = 2.913 J/K

ΔS= 2.913 J/K

5 0
3 years ago
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