Calculate 76.2 to centimeters

How many molecules of co2,h2o,c3h8, and o2 will be present if the reaction goes to completion?

Nadya [2.5K]

Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present

The reaction will be

C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O

so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2

now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O

Hence answer is

molecules of CO2 formed = 6

Molecules of H2O formed = 8

molecules of C3H8 left = 1

molecules of O2 left = 0

7 0

3 years ago

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What is the measured component of the orbital magnetic dipole moment of an electron with (a) ml = and (b) ml = ?

faust18 [17]

ANSWER:

What is the measured component of the orbital magnetic dipole moment of an electron with the values

(a) ml=3

(b ) ml= −4

a) -278 x $10^{-23}$ J/T

b) 3.71 x $10^{-23}$ J/T

STEP-BY-STEP EXPLANATION:

a) ml= 3

Цorb,z = ml Цв = - (3) * (9.27e - 24) = -278 x $10^{-23}$ J/T

b) ml= 3

Цorb,z = ml Цв = - (-4) * (9.27e - 24) = 3.71 x $10^{-23}$ J/T

3 0

3 years ago

list 10 chemical reactions that have benefited your life today. Include the reasons you think each was indeed a chemical reactio

Dennis_Churaev [7]

1) The overall balanced photosynthesis reaction:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.

Plants convert solar energy into the chemical energy of sugars (food).

2) Chemical reaction of cellular respiration:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

Organic carbon (glucose) is transformed into inorganic carbon (carbon dioxide) that goes into atmosphere.

3) Gasoline combustion:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O.

Gasoline is a mixture of many different hydrocarbons: alkanes (paraffins), cycloalkanes and alkenes (olefins).

Gasoline is being burn.

4) Balanced chemical reaction of forming rust: 4Fe + 3O₂ → 2Fe₂O₃.

Forming rust usually last for few months and iron changing in another substance with different properties.

5) An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.

In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

Copper is forming from the solution, that is chemical change.

6) Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.

Making limewater (diluted solution of calcium hydroxide).

7) Neutralization is reaction in which an acid and a base react quantitatively with each other. In this reaction, there are no acids and bases, only salts.

Balanced chemical reaction of vinegar and antacid:

2CH₃COOH(aq) + Mg(OH)₂(aq) → (CH₃COO)₂Mg(s) + 2H₂O(l).

Precipation is formed.

8) Balanced chemical reaction of magnesium and hydrochloric acid:

MgO + 2HCl → MgCl₂ + H₂O(g).

Acid dissolves metal oxides.

9) Chemical reaction of silver in air:

4Ag + 2H₂S + O₂ → 2Ag₂S (tarnish) + 2H₂O.

Silver change color in the air.

10) Forming of patina: 2CuO + CO₂ + H₂O → Cu₂CO₃(OH)₂.

Brass (copper) change color to green.

4 0

3 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

5 0

3 years ago

1 pts

bonufazy [111]

Answer:

secound one is the answer: Distillation

this may help you

5 0

2 years ago

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