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tino4ka555 [31]
2 years ago
14

Write a hypothesis: What is the mole ratio of the cation and the anion in a precipitate?

Chemistry
1 answer:
tankabanditka [31]2 years ago
8 0

Answer:

The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)

Explanation:

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Answer: My opinion is the second one because a lot of humans uses technology and technilogy replaces a lot of stuff.

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3 years ago
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Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta
Vilka [71]

Answer:

147.2g

Explanation:

The full solution can be found in the image attached. Graham's law was applied to the problem. The rate of diffusion of a gas is inversely proportional to its molar mass or vapour density. Molar mass= 2vapour density

8 0
2 years ago
A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235
trapecia [35]

Answer : The value of \Delta E of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}

Molar mass of helium = 4 g/mole

\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole

Now we have to calculate the heat.

Formula used :

q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

c_p = specific heat of helium = 20.8 J/mol.K

T_1 = initial temperature = 310.0 K

T_2 = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

q=nc_p(T_2-T_1)

q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K

q=-19.5J

Now we have top calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 1.21 L

V_2 = final volume = 0.99 L

Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (0.99-1.21)L

w=0.22L.artm=0.22\times 101.3J=22.29J

conversion used : (1 L.atm = 101.3 J)

Now we  have to calculate the value of \Delta E of the gas.

\Delta E=q+w

\Delta E=(-19.5J)+22.29J

\Delta E=2.79J

Therefore, the value of \Delta E of the gas is 2.79 Joules.

3 0
3 years ago
The maximum number of electrons in a valence shell is: <br> a. 2 <br> b. 8 <br> c. 18 <br> d. 32
vodomira [7]
B. 8

Although it may seem that transition metals would have more than 8 valence electrons, they in fact do not.
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You did not link an image of it so I can’t see the rings.
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