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marshall27 [118]
3 years ago
9

Which pair shares the same empirical formula?plz help!

Chemistry
2 answers:
Readme [11.4K]3 years ago
7 0

B

C6H6 = CH

C3H3 = CH

Hope this helps!

n200080 [17]3 years ago
7 0

Answer: B) C_6H_6 and C_3H_3

Explanation: Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

1. The empirical formula for C_2H_4 is CH_2 and C_6H_6 is CH.

2. The empirical formula for C_6H_6 is CH and C_3H_3 is CH.

3. The empirical formula for CH_2 is CH_2 and C_6H_6 is CH.

4. The empirical formula for CH is CH and C_2H_4 is CH_2.

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2 years ago
A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the
Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}

P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

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Which statement is a correct expression of the Law of Conservation of Mass?
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You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1
jeka57 [31]

Answer:

\boxed{\text{36 mL}}

Explanation:

1. Write the balanced chemical equation.

\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O

2. Calculate the moles of HCOOH

\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}

3. Calculate the moles of NaOH.

\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}

4. Calculate the volume of NaOH

c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}

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A catalyst can speed up the rate of a given chemical reaction by
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Decreasing the activation energy needed for the reaction.

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