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Reil [10]
3 years ago
8

How do you unlock a post test on edmentum?

Chemistry
1 answer:
postnew [5]3 years ago
8 0

Explanation:

A post-test or a mastery test can be unlocked if the learner who has the of it previously can complete the given tutorial with the test.

After the tutorial with the test, the learner will have to complete everything and if he passes the test with a completed answer then he will see that the Mastery Test has been unlocked. Also after the mastery test is unlocked it can be done again.

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Joaq...
MrRa [10]

Answer:

all the statements are true of chemical changes

3 0
2 years ago
Which of the following best defines the internal energy of a system?
Anarel [89]
 <span>In thermodynamics, the internal energy of a thermodynamic system, or a body with well-defined boundaries, denoted by U, or sometimes E, is the total of the kinetic energy due to the motion of molecules (translational, rotational, vibrational) and the potential energy associated with the vibrational and electric energy of atoms within molecules or crystals. It includes the energy in all the chemical bonds, and the energy of the free, conduction electrons in metals.</span>
7 0
3 years ago
1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
What is the mass of one mole of carbon-12?
velikii [3]

Answer:

Hello

12 grams

The mass of one mole of carbon-12 atoms is 12 grams.

Hope it helps You.....

Explanation:

4 0
3 years ago
Read 2 more answers
if an object has a mass of 60 grams in a volume of 120 cm3 then calculate the density would this object sink or float
Ivanshal [37]

Answer: 0.5 g/cm^3

Density equals mass divided by volume so..

60/120 is 0.5 g/cm^3

7 0
2 years ago
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