Is the following a properly written thermochemical equation? 2C8H18 + 25O2 → 16CO2 + 18H2O, ΔH =

A binary compound created by reaction of nitrogen and an unknown element E contains 30.46% N and 69.54% E by mass. If the formul

aleksklad [387]

Answer:

15.95

Explanation:

This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.

Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.

In 100 grams of the binary compound we have

30.46 g N

69.54 g E

The number of moles is the mass divided by atomic weight:

mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N

mol E = 65.54 g / A.W E

Thus,

4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18

2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g

So the A.W is 15.94 g/mol which is close the atomic weight of O.

5 0

3 years ago

A partir de que materiales se obtienen los metales? en que se diferencian?

Anon25 [30]

Los metales son materiales que se obtienen a partir de minerales que forman parte de las rocas. Por ejemplo, el metal hierro se extrae de minerales de hierro como la magnetita o la siderita.

6 0

3 years ago

Read 2 more answers

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0 125 kPa

At t = teq 125 - 2x 4x x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$

3 0

3 years ago

A chemist prepares a solution of silver(I) nitrate by measuring out of silver(I) nitrate into a volumetric flask and filling the

leva [86]

Answer:

3.33 M

Explanation:

It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:

" A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits. "

In this case, first we calculate the moles of AgNO₃, using its molecular weight:

85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃

Then we convert the 150 mL of the volumetric flask into L:

150 / 1000 = 0.150 L

Finally we divide the moles by the volume:

0.500 mol AgNO₃ / 0.150 L = 3.33 M

4 0

3 years ago

Suppose there was a release of 1 mole of Alpha emission particle and 1 mole of Beta emission particles and both particles are ac

Vitek1552 [10]

Answer:

DT the best and they I'm on to work use as some help new thread so this communication contains you would appreciate everything about getting started using some new version from here or email the day car rental house has come home today was automatically and your company based website has anyone that a big hugs bhi to

Explanation:

DT the best and they I'm on to work use as some help new thread so this communication contains you would appreciate everything about getting started using some new version from here or email the day car rental house has come home today was automatically and your company based website has anyone that a big hugs bhi to

6 0

3 years ago

Is the following a properly written thermochemical equation? 2C8H18 + 25O2 → 16CO2 + 18H2O, ΔH = –5,471 kJ/mol