All categories

3 years ago

Consider the following intermediate reactions.

Chemistry

2 answers:

Alja [10]3 years ago

4 0

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

schepotkina [342]3 years ago

3 0

Answer:

The answer is A on edge

Explanation:

You might be interested in

A star is mainly made of hydrogen gas, but most classes of stars include other elements such as helium, calcium, and other heavi

Xelga [282]

That would be true, since most other stats include helium, calcium, etc

6 0

2 years ago

Read 2 more answers

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

8 0

3 years ago

Draw one representation that shows the intermolecular interactions between NH_3 and water and another that shows the intermolecu

Lera25 [3.4K]

Answer:

NH3 has greater water solubility due to intermoleculate interactions

Explanation:

Hi:

If we represent the structures of NH3 and SbH3 we can see that they are similar to the naked eye, this is because N and Sb belong to the same group of the periodic table (group 15).

However, the electronegativity of N is greater than that of Sb. The NH3 molecule is polar and can form an intermolecular interaction called hydrogen bridge with water.

Sb is less electronegative than N. The SBH3 molecule forms an intermolecular interaction with water called dipole-induced dipole.

The zone with positive charge density of the water molecule (hydrogens) is oriented towards the zone with positive charge density of SBH3 (the pair of electrons not shared)

Stronger intermolecular junctions allow greater solubility of NH3 molecules.

Successes in your homework

Download pdf

5 0

3 years ago

Describe Write your own caption for this photo.

koban [17]

Answer:

Sit by the fire to warm up

Explanation:

5 0

2 years ago

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.

See figure 1

I hope it helps!

4 0

3 years ago