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2 years ago

Part A: Calculate the mass of butane needed to produce 75.6g of carbon dioxide.

Chemistry

1 answer:

lisabon 2012 [21]2 years ago

5 0

Answer:

Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.

Explanation:

Part B:

The mass of water produced when 4.86 g of butane(C4H10) react with excess oxygen is calculated as below

calculate the moles of C4H10 used = mass/molar mass

moles = 4.86g/58 g/mol =0.0838 moles

write a balanced equation for reaction

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

by use of mole ratio between C4H10 to H2O which is 2:10 the moles of

H20= 0.0838 x10/2 = 0.419 moles of H2O

mass = moles x molar mass

=0.419 molx 18 g/mol = 7.542 grams of water is formed

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Match the following chemical reactions:

S_A_V [24]

Answer:

Synthesis - 4

reversible- 2

exchange- 1

decomposition-3

Explanation:

In synthesis reaction two or more components combines to form a single product. example 2H2+O2⇒2H2O

In reversible reaction two reactants combine to form two products . The products then reacts and forms back the reactants. example N2 +3H2 ⇒2NH3

In exchange reaction there is an alternation of ions of reactants to form new products. AB+CD ⇒AC + BD

In decomposition reaction, molecules of a compound break down by the action of heat or light or catalyst. example CaCO3 ⇒CaO +CO2

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3 years ago

What is the mass of 5 moles of Fe2(CO3)3 ?

sineoko [7]

Answer:

1218.585

Explanation:

Looking at the subscripts we know there are 2 atoms of Fe, 3 atoms of C, and 6 of O.

Take the molar mass of each atom (from the periodic table) and multiply by the # of atoms

Fe: 55.845×2= 111.69

C: 12.011×3= 36.033

O:15.999×6=95.994

Add the values together: 243.717 g/mol

That is 1 mole of the molecule. Multiply by 5 for the final answer.

243.717×5=1218.585

7 0

3 years ago

Compared to the nucleus 5626fe, what is the density of the nucleus 112 48cd?

tia_tia [17]

For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.

n(Cd) / n(Fe)=1

<h3>What is the density of the nucleus 112 48cd?</h3>

Generally, the equation for the density is mathematically given as

d=\frac{A}{4/3}\piR^3

Therefore

n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3

n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3

n(Cd) / n(Fe)=1

In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same

Read more about density

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2 years ago

Which of the following pairs contains a conjugate acid-base pair?

zzz [600]

Remember that a conjugate acid-base pair will differ only by one proton.
None of the options you listed are conjugate acid-base pairs as none of them differ only by one proton (or H⁺)
An example of a conjugate acid-base pair would be NH₃ and NH₄⁺NH₃ + H₂O --> NH₄⁺ + OH⁻NH3 is the base, and NH₄⁺ is the conjugate acid

5 0

3 years ago

Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made

arlik [135]

Answer: The empirical formula is $C_3H_3O$ .

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 12.24 g

Mass of $H_2O$ = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, = $\frac{12}{44}\times 12.24=3.338g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, = $\frac{2}{18}\times 2.505=0.278g$ of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.278}{0.104}=3$

For H = $\frac{0.278}{0.104}=3$

For O = $\frac{0.104}{0.104}=1$

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is $C_3H_3O$ .

5 0

3 years ago