All categories

3 years ago

128.3 grams of methane (CH4) is equal to how many moles of methane

2 answers:

6 0

About 8.0 moles of methane.Number of moles = MassMolar mass.

And thus we get the quotient:

128.3⋅g16.04⋅g⋅mol−1=8.0⋅moles of methane.

Note that the expression is dimensionally consistent, we wanted an answer in moles, and the quotients gives, 1mol−1=11mol=mol as required.

faltersainse [42]3 years ago

4 0

Answer : The moles of methane are, 8.01875 moles

Explanation : Given,

Mass of $CH_4$ = 128.3 g

Molar mass of $CH_4$ = 16 g/mole

Now we have to calculate the moles of methane.

Formula used :

$\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}$

Now put all the given values in this formula, we get the moles of methane.

$\text{Moles of }CH_4=\frac{128.3g}{16g/mole}=8.01875moles$

Therefore, the moles of methane are, 8.01875 moles

You might be interested in

Which of the following is a good conductor of electricity?

maks197457 [2]

Answer:

Explanation:

Iron is the correct answer

3 0

2 years ago

Read 2 more answers

How do you find a theoretical mass? Is there a difference between theoretical mass and theoretical yield?

swat32

Theoretical yield is the quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction. It is the amount of product resulting from a perfect chemical reaction and thus not the same as the amount you'll actually get from a reaction.

3 0

3 years ago

Why was the line of best fit method used to determine the experimental value of absolute zero?

BartSMP [9]

Answer:

Because the x-intercet of the graph represents volume zero, which indicates the minimum possible temperature or absolute zero.

Explanation:

Charle's Law for ideal gases states that, at constant pressure, the temperature and the volume of a sample of gas are protortional.

$\dfrac{V}{T}=k\\\\V=kT$

That means that the graph of the relationship between Temperature, in Kelivn, and Volume is a line, which passes through the origin.

When you work with Temperature in Celsius, and the temperature is placed on the x-axis, the line is shifted to the left 273.15ºC.

Meaning that the Volume at 273.15ºC is zero.

You cannot reach such low temperatures in an experiment, and also, volume zero is not real.

Nevertheless, you can draw the line of best fit and extend it until the x-axis (corresponding to a theoretical volume equal to zero), and read the corresponding temperature.

Subject to the experimental errors, and the fact that the real gases are not ideal, the temperature that you read on the x-axis is the minimum possible temperature (absolute zero) as the minimum possible volume is zero.

3 0

3 years ago

ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a

Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{20\text{ min}}$

$k=3.465\times 10^{-2}\text{ min}^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.465\times 10^{-2}\text{ min}^{-1}$

t = time passed by the sample = 80 min

a = initial amount of the reactant = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}$

$a-x=0.100M$

Therefore, the concentration of A after 80 min is, 0.100 M

3 0

2 years ago

To show the electron configuration for an atom, when would it be better to use an orbital notation than to use a written configu

anzhelika [568]

I believe it would be better to use an orbital designation than the written configuration, if the number of electrons in the ground state of the atom are quite high for the given element, as above 50, for instance.

This saves space and also one can see the discrete quantized energy levels associated with the subshells of the main energy levels if written in orbital designation.

7 0

3 years ago

Read 2 more answers