All categories

3 years ago

A gas has an empirical formula CH4. 0.16g of the gas occupies a volume of 240cm^3 What is the molecular formula of the gas???.

Chemistry

1 answer:

yulyashka [42]3 years ago

8 0

Explanation and answer:

The molar volume of a gas is 22.4 L at 0 deg. C.

So the molar mass of the gas is, by proportion,

0.16 g * (22400 mL)/(240 mL) = 14.93 g

The molar mass of (CH4)n = 12+4(1) = 16.

So n = 15.93/16 = 1, or the molecular formula is CH4.

Note: The temperature at which the volume was observed was not given. If 240 cm^3 was observed at 20 deg.C, then the volume at 0 deg.C would be

V=240*(273+0)/(273+20) = 223.6

The molar mass = 0.16*22400/(223.6) = 16.03

which gives n = 16/16.03 = 1 again, but more accurately.

You might be interested in

Lipids that are liquid at room temperature are known as ____.

zaharov [31]

Lipids that are liquid at room temperature are known as oils.

7 0

3 years ago

What two atoms are held together by a covalent bond in NaHCO3

gtnhenbr [62]

Answer: covalent bond is the force of attraction that holds together two atoms that share a pair of valence electrons. Covalent bonds form only between atoms of nonmetals. The two atoms that are held together in a covalent bond may be atoms of the same element or different elements.

Explanation:

8 0

3 years ago

Describe two physical and two chemical changes involved in cooking. (4 points)

Ivenika [448]

Physical changes are when things get changed without altering chemical consistencies, which is melting solid butter into liquid one, or boiling water. Chemical changes are things such as caramelizing sugar when making sweets, or when carbon dioxide is created and released when baking bread.

7 0

4 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

4 years ago

Why are the oxidation and reduction half-reactions separated in an<br> electrochemical cell?

Mrac [35]

Answer:

The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.

Explanation:

7 0

3 years ago

Read 2 more answers