Answer:
Decomposition of potassium chlorate yields potassium chloride and oxygen as:
2KClO
3
→2KCl+3O
2
Thus 2 moles of potassium chlorate yields 3 moles of oxygen gas.
2 moles of potassium chlorate =2×122.5=245g of potassium chlorate
At STP, the volume occupied by 1 mol of gas =22.4 dm
3
the volume occupied by three moles of a gas =3×22.4=67.2dm
3
Therefore, 245g of potassium chlorate yields 67.2dm
3
of oxygen gas
To liberate 6.72 dm
3
oxygen amount of potassium chlorate required is
=
67.2
245
×6.72=24.5g
Hence, 24.5 g of potassium chlorate required to liberate 6.72 dm
3
of oxygen at STP
Answer:
The correct answer is 23,4375 grams of C.
Explanation:
First, we will identify the limiting reagent, which is the one consumed first. Since each gram of B consumes two grams of A, the product to be consumed first is A.
We calculate the amount of A and B that is consumed in the first 10 minutes is 5 grams of B and 10 grams of A.
Now we calculate the constant that relates the speed of reaction to the amount of reagent in each component.
20 grams of B and 40 grams of A will react completely given the limiting reagent A. We calculate the constant that relates the reaction speed (grams of product/time) with the instantaneous amounts of unconverted A and B. We calculate the constant that relates the reaction speed (grams of product/time) with the instantaneous amounts of unconverted A and B. We calculate the constant that relates the reaction speed (grams of product/time) with the instantaneous amounts of unconverted A and B.
So:
15g/10min = k*(20g*40g)
k=1.5 g/min / 800 g^2
k=1,875 *10^-3 g^3 /min
Now, we calculate the amount of reagent that forms when 20 minutes pass:
x / 10 min = 1,875 *10^-3 g^3/min * (15g * 30g)
x= 8,4375 grams
So we can say that after 20 minutes, 23.4375 grams of C will form.
Have a nice day!
Answer:
The oxidation number of an atom is the charge it appears to have when you count the electrons according to some arbitrary rules. The oxidation number of an atom depends on the other atoms in the substance.
Explanation:
For example, In KCl, the oxidation number of Cl is 0.
If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.
An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Given :
The ideal gas equation is given below.
n = PV/RT
n = 86126.25 x 0.0024 / 8.314 x 287
n = 0.622 / molar mass (n = Avogardos number)
Molar mass = 7.18 g
Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g
More about the ideal gas equation link is given below.
brainly.com/question/4147359
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B) they have no charges and are inside an atom.*
