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2 years ago

Write skeleton equations for these reactions. a. butane(l) + oxygen (g) --> carbon dioxide(g) + water(l).

1 answer:

4 0

Skeleton equation is actually a chemical reaction whee the reactants and products are written in their chemical formula. In this problem, this is actually a combustion reaction of butane. The chemical formula for butane is C₄H₁₀.

C₄H₁₀ (l) + 6.5 O₂ (g) --> 4 CO₂ (g) + 5 H₂O (l)

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Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-1

Irina-Kira [14]

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

$m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}$

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:

$1 = \%_{16} + \%_{17} + \%_{18}$

$1 = x + 3.2 \cdot 10^{-4} + \%_{18}$

$\%_{18} = 1 - x - 3.2 \cdot 10^{-4}$

Hence, the percent abundance of O-18 is:

$m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}$

$15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)$

$x = 0.980614 \times 100 = 98.0614 \%$

Hence, the percent abundance of oxygen-18 is:

$\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%$

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!

8 0

2 years ago

What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3

Len [333]

Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
M acid= 0.129 M
V base= ?
M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

$V base=\frac{0.0137 L*0.129 M}{0.135 M}$

V base=0.0131 L = 13.1 mL

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

4 0

3 years ago

Cyclobutane simplified structural formula please

daser333 [38]

Answer:

Cyclobutane is a cycloalkane and organic compound with the formula (CH2)4.

Explanation:

5 0

2 years ago

A plane mirror is useful for seeing:

AURORKA [14]

C. A nearly identical image.

6 0

2 years ago

Name the organic molecule

tekilochka [14]

Answer:

hydrocarbons are the answer

6 0

2 years ago