If gas is truly to be classified as a form of matter this means it can be measured. What form of gas can be measured?

Chemistry

2 answers:

N76 [4]2 years ago

6 0

Answer:

A dry gas, that is a natural gas, is the form in which gas can be measured.

And gas is measured in cubic meter, that is, it's volume is measured and the size of gas can be determined.

All the best!

juin [17]2 years ago

3 0

Answer:

Gas is sometimes measured in cubic feet at a temperature of 60 degrees Fahrenheit and an atmospheric pressure of 14.7 pounds per square inch. Gas production from wells is discussed in terms of thousands or millions of cubic feet (Mcf and MMcf). Resources and reserves are calculated in trillions of cubic feet (Tcf).

You might be interested in

Urgent! which element do you think has a greater mass? cadmium or zinc?

Alex Ar [27]

Answer:

zinc

Explanation:

5 0

2 years ago

Read 2 more answers

State which of the following set of quantum number would be possible and which would be permissible for an electron in an atom.

raketka [301]

Answer:

The correct option is: e) n=4, l=3, $m_{l}$ = - 2 , $m_{s}$ = - 1/2

Explanation:

The four quantum numbers that describe the electrons in an atom are: The principal: n, azimuthal: l, magnetic: m, and spin: s.

For a given electron, the values of the quantum numbers should be in the given range- principal quantum number: n ≥ 1;

azimuthal quantum number: 0 ≤ l ≤ n − 1;

magnetic quantum number: −l ≤ $m_{l}$ ≤ +l;

spin quantum number: −s ≤ $m_{s}$ ≤ +s

Now, for n= 4,

The value of l: 0 to n − 1 = 0 to 3 = 0, 1, 2, 3

The value of $m_{l}$ for l= 3 : -3 to +3 = -3, -2, -1, 0, +1, +2, +3

The value of $m_{s}$ : -1/2, +1/2

Therefore, the given set of quantum numbers: n=4, l=3, $m_{l}$ = - 2 , $m_{s}$ = - 1/2; are possible and permissible for an electron in an atom.

5 0

3 years ago

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Romashka-Z-Leto [24]

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we want to find $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?$ when the height is 2 cm.