Match each molecular formula to the corresponding empirical formula. 1) C₄H₈O₂ 2) C₄H₁₂O₄ 3) C₂H₁₀O₂ A) CH₃O B) CH₅O C) C₂H₄O

Answers

Answer 1

Answer: So empirical formulas would be the most simplified version where the proportions of the elements are the most simplified version. The best way to think of this would be like a recipe. If you want to make one cake and it says use two eggs then when you want two cakes you use four eggs. Each element works similarly. 1) divide by two as the greatest common factor (GCF) between 4,8, and 2 is 2 making 1) simplify to C2H4O so 1) C4H8O2 is the molecular formula of C)C2H4O. Then for 2) C4H10O4 we think of the GCF again. Biggest factor or GCF between 4,12,and 4 is 4, so divide the numbers in C4H12O4 by 4 making CH3O which is choice A) CH3O making 2) the molecular formula of A). Lastly do the same thing but from process of elongation 3) would be the molecular formula of B) CH5O. :) hope this helps :)))

Answer 2

Answer:

Final answer:

The molecular formula C4H8O2 corresponds to the empirical formula C2H4O, C4H12O4 corresponds to CH3O, and C2H10O2 corresponds to C2H5O.

Explanation:

In chemistry, an empirical formula represents the simplest whole-number ratio of atoms in a compound, while a molecular formula represents the actual number of atoms in a molecule of the compound. To match each molecular formula to the corresponding empirical formula, divide the subscripts in the molecular formula by their greatest common factor. Hence:

C4H12O4 matches with B) CH3O
C2H10O2 fits with C) C2H5O

Learn more about Empirical and Molecular Formulas here:

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The reaction A→B has been experimentally determined to be second order. The initial rate is 0.0100M/s at an initial concentration of A of 0.300M. Determine the initial rate at [A]=0.900M.

Answers

Answer:

0.09 M/s is the initial rate when concentration of reactant A is 0.900 M.

Explanation:

A → B

1) Initial concentration of A = [A] =0.300 M

Rate constant of the reaction , k = ?

Rate equation for the second order kinetic is given as:

$R=k[A]^2$

$0.0100M/s=k[0.300 M]^2$

$k=(0.0100M/s)/([0.300 M]^2)=0.1111 M^(-1) s^(-1)$

2) Rate of the reaction when initial concentration of A was 0.900M be R'.

[A] = 0.900M

$R'=k[A]^2$

$R'=0.1111 M^(-1) s^(-1)* [0.900 M]^2$

$R'=0.09 M/s$

What element is used in a liquid metal used in thermometers

Answers

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The amount of heat energy required to raise the temperature of a unit mass of a material one degree isA. absolute zero.
B. a Joule.
C. its heat capacity.
D. a Btu.

Answers

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The percentage by mass of Br in the compoundAlBr3 is closest to
(1) 10.% (3) 75%
(2) 25% (4) 90.%

Answers

Answer: The correct answer is option(4).

Explanation:

Molar mass of the aluminium bromide, $AlBr_3=266.69 g/mol$

Atomic mass of bromine = 79.90 g/mol

Percentage of Mass of bromine in $AlBr_3$ :

$\% Br\text{ by mass}=\frac{\text{Atomic mass of Br}* \text{Number of Br atoms}* 100}{\text{Molar mass of }AlBr_3}$

$\% Br\text{ by mass}=(79.90 g/mol* 3* 100)/(266.69 g/mol)=89.97\%\approx 90\%$

Hence, the correct answer is option(4).

What you have to do it find the percentage / solve it. Divide the mass by the entire mass.
239.7 ÷ 266.68 = 0.898 × 100 = 90% Br.

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Answers

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Answers

Answer:

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Explanation:

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