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FrozenT [24]
3 years ago
5

Starting with the unbalanced equation for the combustion of ethanol, C2H5OH(l)+O2(g)→CO2(g)+?H2O(l) what coefficient should be p

laced in front of H2O to balance the hydrogen atoms?
Chemistry
1 answer:
lianna [129]3 years ago
6 0

Answer:

C2H5OH(l) + O2(g) → CO2(g) + ?H2O(l)

The coefficient should be 3

Explanation:

C2H5OH(l) + O2(g) → CO2(g) + ?H2O(l)

The chemical equation is a reaction between ethanol and oxygen . The reaction yield is carbon dioxide and water. Balancing the chemical equation requires one to make the number of atoms on the reactant side be equal to the number of atoms on the product side.

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The hydrogen atom on the left hand side of the equation is 6. Therefore, the number of atom of hydrogen should be 6 on the right hand side. The number of oxygen atom on the left hand side is 7. Therefore, the number of Oxygen atom on the right hand side should  7 . Oxygen atom in carbon dioxide(right hand side) is 4 atoms, therefore, it requires 3 atoms of oxygen to make the  oxygen atoms 7.

Using 3 as the coefficient of water in the equation will make hydrogen 6 and oxygen 3 .Adding the 3 oxygen to the already 4 atoms of oxygen in carbon dioxide will make oxygen 7 atoms

 

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The answer is C. 299k
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Compute the values. Express these answers to the hundredths place (i.e., two digits after the decimal point). log ( 2.1 ) = ln (
zepelin [54]

Answer:

x = 4.17

y = 1.86

Explanation:

0.62 = log(x)

x = 10^0.62 = 4.17 ( to the nearest hundredth)

0.62 = ln(y)

y = e^0.62 = 1.86 (to the nearest hundredth)

8 0
3 years ago
The nucleus of an atom ok K-42 contains
horsena [70]
Number of proton K=19

so, 42 - 19 =23

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3 0
3 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
a scuba diver's tank contains 0.29g of oxygen compressed into volume of 2.3L. What is the pressure in the tank at 9C?
viva [34]

Answer:

The answer to your question is P = 0.18 atm

Explanation:

Data

mass of O₂ = 0.29 g

Volume = 2.3 l

Pressure = ?

Temperature = 9°C

constant of ideal gases = 0.082 atm l/mol°K

Process

1.- Convert the mass of O₂ to moles

                16 g of O₂ -------------------- 1 mol

                0.29 g of O₂ ----------------   x

                    x = (0.29 x 1)/16

                    x = 0.29/16

                    x = 0.018 moles

2.- Convert the temperature to °K

Temperature = 9 + 273 = 282°K

3.- Use the ideal gas law ro find the answer

              PV = nRT

-Solve for P

              P = nRT/V

-Substitution

              P = (0.018 x 0.082 x 282) / 2.3

-Simplification

              P = 0.416/2.3

-Result

               P = 0.18 atm

5 0
3 years ago
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