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Kobotan [32]
3 years ago
11

Doing Science plz help

Chemistry
2 answers:
tekilochka [14]3 years ago
6 0
The answer is C
Because it’s balanced you know this because of law conversion of mass.
Murljashka [212]3 years ago
4 0

Answer:

My guess is b or c but its robably wrong

Explanation:

I just also need points sorry <3

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1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data
earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v =  1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15  K

v₂ = 0.773 T₂ = 273.15  K

v₃ = 1.057 T₃ = 373.15  K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533  = 0.00284×T - 0.001167

v =  0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 =  0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411  -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

3 0
3 years ago
Why doesn’t the moon move away from earth
Leni [432]
The earth has the moon captured in its gravity. this keeps the moon in an elliptical orbit
8 0
2 years ago
What is the pH of a solution that has a pOH of 5.9
Stella [2.4K]

Answer:

pH = 8.1

Explanation:

Assuming that we are at 25 degrees Celsius, pH + pOH = 14.

We can then plug in the given pOH and solve for pH:

pH + pOH = 14

pH + 5.9 = 14

pH = 14 - 5.9 = 8.1

7 0
3 years ago
some components of ink are minimally attracted to the stationary phase and very soluble in the solvent. where are these componen
Trava [24]
They’ll have moved the farthest, since the solvent is best at carrying those kinds of materials.
5 0
2 years ago
Read 2 more answers
n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form
Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

pK_a=8.0

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}

Now put all the given values in this expression, we get:

6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}

\frac{[Deprotonated]}{[Protonated]}=0.01  

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

\frac{[Protonated]}{[Deprotonated]}=100  

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0
2 years ago
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