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3 years ago

15

Please help PLSSSS 2020020200

1 answer:

Marat540 [252]3 years ago

5 0

Answer:Well the water molecules attract each other, and the oil molecules stick together.

sorry number two im not sure sorry:(

Explanation:

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A sample of gas is at 273°K and 2.00 atm with a volume of 6.00 liters. When the gas is compressed to 3.00 liters and the pressur

horrorfan [7]

Answer: 0.00488K

Explanation: using the general gas equation

(P1V1)/T1 =(P2V2)/T1

Substitute

2*6/273= 3*3/T

Simplify

T= 0.00488°K

3 0

4 years ago

Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Us

madam [21]

Answer:

$d=0.92\frac{kg}{m^{3}}$

Explanation:

Using the Ideal Gas Law we have $PV=nRT$ and the number of moles n could be expressed as $n=\frac{m}{M}$ , where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

$PV=\frac{m}{M}RT$

If we pass the V to divide:

$P=\frac{m}{V}\frac{RT}{M}$

As the density is expressed as $d=\frac{m}{V}$ , we have:

$P=d\frac{RT}{M}$

Solving for the density:

$d=\frac{PM}{RT}$

Then we need to convert the units to the S.I.:

$T=100^{o}C+273.15$

$T=373.15K$

$P=1bar*\frac{0.98atm}{1bar}$

$P=0.98atm$

$M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}$

$M=0.0289\frac{kg}{mol}$

Finally we replace the values:

$d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}$

$d=9.2*10^{-4}\frac{kg}{L}$

$d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}$

$d=0.92\frac{kg}{m^{3}}$

5 0

3 years ago

the water in a tea pot is heated on a stove top. The temperature of the water increases. Is this an endothermic or exothermic pr

marusya05 [52]

It is an endothermic process

5 0

3 years ago

How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.

Slav-nsk [51]

Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

7 0

3 years ago

All charges in a compound must have ?

MakcuM [25]

They must have both cations and anions.

I hope this helps.

8 0

3 years ago