All categories

3 years ago

How many moles of N are in 0.241 g of N2O?

Chemistry

2 answers:

gtnhenbr [62]3 years ago

6 0

0.00547 moles of N.

eduard3 years ago

6 0

Answer:

Explanation:

Calculate the molar mass of the $N_{2}O$ :

Molar mass N = $14\frac{g}{mol}$

Molar mass O = $16\frac{g}{mol}$

Molar mass $N_{2}O$ = 2*(Molar mass of N) + Molar mass of O

Molar mass $N_{2}O$ = $2*(14\frac{g}{mol})+16\frac{g}{mol}$

Molar mass $N_{2}O$ = $44\frac{g}{mol}$

Calculate the number of moles of N in 0.241g of $N_{2}O$ :

$0.241gN_{2}O*\frac{1molN_{2}O}{44gN_{2}O}*\frac{2molesN}{1molN_{2}O}=0.01molesN$

You might be interested in

What is the net force f⃗ net that the dipole experiences due to the electric field?

Talja [164]

In physics, net force is the overall force acting on an object. In order to calculate the net force, the body is isolated and interactions with the environment or other constraints are represented as forces and torques in a free-body diagram.

3 0

3 years ago

What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?

Brut [27]

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams

7 0

3 years ago

As the Amplitude of the wave decreases what be does the frequency do

Alexxx [7]

Answer: pic of the test

Explanation:

4 0

2 years ago

Write the concentration equilibrium constant expression for this reaction.

Akimi4 [234]

In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.

The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)

The equilibrium constant = product of concentration of products / product of concentration of reactants

(Here, H2O is not considered as its concentration is very high)

So, Keq = [CN⁻] / [HCN] [OH⁻]

8 0

3 years ago

Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802

Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo ( $Q$ ), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto ( $\bar {Q}$ ), en kilojoules por mol, dividido por su masa molar ( $M$ ), en gramos por mol: