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4 years ago

How many total oxygen (O) atoms are present in the products side of the equation C2H5OH + 3O2 → 2CO2 + 3H2O?

Chemistry

1 answer:

kifflom [539]4 years ago

3 0

Answer:

There are seven oxygen atoms in total on the product side.

Explanation:

The oxygen atom is present in both the species of CO₂ and H₂O. As there are two molecules of CO₂, so the number of oxygen atoms from it will be 4. There are three molecules of H₂O in the product side so oxygen atoms from it will be three. So, the total number of oxygen from both species is seven.

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Please help with the attached picture question 21, 22, 23

klasskru [66]

Explanation:

21. The given molecule for cracking is tetradecane.

On cracking it forms one mole of decane (C10H22) and two moles of ethene gas.

The chemical equation is shown below:

$C_1_4H_3_0->C_1_0H_2_2+2C_2H_4$

22. The essential condition for the formation of an ester is the reaction of alcohol and acid in presence of concentrated sulfuric acid.

Thus among the given options, the first option is the correct one.

23. Isomers of butanol are shown below:

It is 2-butanol.

The position of -OH group changes to the second carbon.

5 0

3 years ago

A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10

lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0

3 years ago

Which of these bonds is the weakest?

Olenka [21]

Answer:

The first one has the weakest bond

Explanation:

The number of atoms together shows how strong the bonds in the second bond it is 4 bonds, the third is 8 bonds, the 4th one is 4 bonds thus why the first bond is the weakest.

6 0

3 years ago

Which of the following describes the process of deflation?

goldenfox [79]

The answer is C. <span>The wind removes surface particles, resulting in lower land surface.

Deflation is the process of lifting and moving of tiny particles of soil or sand, brought about by the wind. It can also be associated with soil erosion. This is a common phenomenon in deserts and sand dune areas near the coast.</span>

4 0

3 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

5 0

2 years ago