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Sliva [168]
2 years ago
14

Week 4 learning task 1 science 4th quarter​

Chemistry
1 answer:
cestrela7 [59]2 years ago
8 0

Answer:

what???

Explanation:

what??????????

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What is the mass percentage of carbon in 5.000 G of sucrose
Crank
17) 8.4 / 20 x 100

18) 20 . 0.5150

19) 6,50% because (as you said) the law of definite proportions states that regardless of the amount, a compound is always composed of the same elements in the same proportion by mass
3 0
3 years ago
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A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press
earnstyle [38]

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)  

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)  

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg

6 0
3 years ago
What causes one water molecule to stick to another water molecule?
stiks02 [169]

Answer:

Hydrogen Bonds Make Water Sticky

Explanation:

This sticking together of like substances is called cohesion. Depending on how attracted molecules of the same substance are to one another, the substance will be more or less cohesive. Hydrogen bonds cause water to be exceptionally attracted to each other. Therefore, water is very cohesive.

3 0
3 years ago
For the titration of a weak acid with a strong base what is the pKa of the weak acid if the pH is 6.72 at the equilvalence point
Paladinen [302]

Explanation:

We have to calculate pK_{a} value.

It is known that at the equivalence point concentration of acid is equal to the concentration of anion formed.

Hence,         [HA] = [A^{-}]

Now, relation between pK_{a} and pH is as follows.

                   pH = pK_{a} + log \frac{[A^{-}]}{[HA]}

Putting the values into the above formula as follows.

                   pH = pK_{a} + log \frac{[A^{-}]}{[HA]}

                  4.23 = pK_{a} + log (1)            (as [HA] = [A^{-}])

                   pK_{a} = 4.23                (as log (1) = 0)

or,                   pK_{a} = 4

Thus, we can conclude that pK_{a} of given weak acid is 4.  

                   

6 0
3 years ago
What is the heat of combustion of ethane, c2h6, in kilojoules per mole of ethane? Enthalpy of formation values can be found in t
Andre45 [30]

The balanced combustion reaction for one mole of ethane will be

C2H6 + 3.5 O2  ---> 2CO2(g) + 3H2O (l)

Enthalpy of reaction of combustion = ∑ΔHf (products)- ∑ΔHf (reactant)

ΔHf = enthalpy of formation

The enthalpy of formation of oxygen = 0 as it is an element in its native state

ΔHf C2H6 = -83.820 kJ / mole

ΔHf H2O = -285.830 kJ / mole

ΔHf CO2 = -393.509 kJ / mole

ΔHrxn = [2 X ΔHf CO2 + 3 X ΔHf H2O] - [ΔHf C2H6]

ΔHrxn = [2 X (-393.509) + 3 (-285.830)] - [-83.820]

ΔHrxn = -1560.688 kJ / mole

7 0
3 years ago
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