All categories

2 years ago

How many molecules are in 0.25 grams of dinitrogen pentoxide?

Chemistry

2 answers:

ZanzabumX [31]2 years ago

7 0

Answer: $0.014\times 10^{23}molecules$

Explanation: Moles can be calculated by using the formula:

$\text{no of moles}=\frac{\text{given mass}}{\text{Molecular mass}}$

$\text{no of moles}=\frac{0.25g}{108g/mol}=0.002moles$

1 mole of $N_2O_5$ contains = $6.023\times 10^{23}molecules$

0.002 mole of $N_2O_5$ contains = $\frac{6.023\times 10^{23}}{1}\times 0.002=0.014\times 10^{23}molecules$

Vesna [10]2 years ago

4 0

In order to get the answer, you have to know the formula of how to convert grams to mole and by that you can get the number of molecules.
The answer is 2 moles of N2O5. This is shown by the solution:
(0.25 g N2O5) (1 mol/ 108 g)=2.31 molecules
The answer is 2 molecules.

You might be interested in

5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is

GenaCL600 [577]

1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V = 7.9 atm

3 0

2 years ago

Read 2 more answers

Compound X has a molar mass of 104.01·gmol−1 and the following composition: element mass % nitrogen 26.93% fluorine 73.07% Write

Stolb23 [73]

Answer:

Explanation:

شخبارج فاطمه

4 0

2 years ago

Read 2 more answers

. Un cuerpo irregular se introduce dentro de una probeta que inicialmente tiene un volumen de 20 ml como muestra la figura Despu

MrMuchimi

Answer:

El volumen del cuerpo es el mismo al comienzo de la experiencia.

Explicación:

El volumen del cuerpo es el mismo al principio porque el volumen no cambia si la temperatura permanece igual. Si cambiamos la temperatura i. mi. Al aumentar la temperatura, las moléculas comienzan a expandirse y se produce un aumento de volumen mientras que cuando disminuimos la temperatura, las moléculas de esa sustancia comienzan a contraerse y el volumen de esa sustancia disminuye. Entonces concluimos que el volumen depende de la temperatura.

6 0

2 years ago

A liquid with a high viscosity cannot be a mixture.

enot [183]

It can be made true by changing "cannot" to "can".

5 0

2 years ago

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$