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Temka [501]
3 years ago
9

1.Adipic acid contains 49.32 % Carbon, 43.84 % Oxygen, and 6.85 % hydrogen by mass. What is the Empirical Formula of Adipic acid

?
2.Find the Empirical Formula of a compound which contains 54.93 % Potassium, 38.73% Boron and 6.34 % hydrogen.
3.Find the Empirical Formula for a compound which contains 26.8 % Tin, 16 % Chlorine and 57.2% Iodine.
4.What is empirical formula of a compound that 41.38% Magnesium, 55.17% Oxygen, and 3.45% Hydrogen
5.Find the empirical formula for a compound which contains 32.8% chromium and 67.2% Chlorine.
Chemistry
1 answer:
IceJOKER [234]3 years ago
6 0
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The compounds in alcohol that contribute to hangover symptoms are called __________.
Advocard [28]

Answer: Ethanol

Explanation: When we consume alcohol (ethanol), the ethanol has a wide range of effects on the body and it also metabolize in the liver producing acetaldehyde which is believed to be a main cause of hangover

5 0
3 years ago
The area in which certain types of plants or animals can be found living in close proximity to each other is called a_________
posledela
I think the answer to this is A
5 0
4 years ago
Read 2 more answers
1. Classify the following matter as a pure substance or a mixture.<br> Potassium
dlinn [17]

Pure substance, basically, it’s on the periodic table.

4 0
3 years ago
What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 mole of c
yuradex [85]

Answer:

Theoretical moles of V are 1.6 moles

Explanation:

The theoretical yield of a reaction is defined as the amount of product you would make if all of the limiting reactant was converted into product.

In the reaction:

V2O5(s) + 5Ca(i) → 2V(i) + 5CaO(s)

Based on the reaction, 1 mol of V2O5 needs 5 moles of Ca for a complete reaction. As there are just 4 moles, <em>limiting reactant is Ca.</em> As there are produced 2 moles of V per 5mol of Ca, Theoretical moles of V are:

4 moles of Ca × (2mol V / 5Ca) = <em>1.6 moles of V</em>

<em></em>

I hope it helps!

5 0
3 years ago
For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in
Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>

<em>C(s) + 2H2(g) <--> CH4(g). </em>

<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2)²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78)²

<u><em>PpCH4 = 1.44 atm</em></u>

6 0
3 years ago
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