CHEMISTRY MIDDLE SCHOOL

Which formula is an empirical formula?A. N2O4

B. NH3

C. C3H6

D. P4O10

Answers

Answer 1

Answer:

The compound that represents empiricalformula is NH3.

EMPIRICAL FORMULA:

Empirical formula refers to formula of a compound that depicts the simplest whole number ratio of its constituent element.

The empirical ratio of the elements of a compound should not be able to be broken down further. For example: compounds such as N2O4, C3H6, and P4O10 can be further broken down into NO2, CH2 and P2O5.

This means that NH3 is the empiricalformula because it is the simplest whole number ratio of the compound.

Learn more at: brainly.com/question/21280037?referrer=searchResults

Answer 2

Answer:

Hello!

The answer is:

The empirical formula is the option B. $NH_(3)$

Why?

The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

$MolecularFormula=n(EmpiricalFormula)$

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.

Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)

A. $N_(2)O_(4)$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$N_(2)O_(4)=2(NO_(2))$

B. $NH_(3)$

It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.

C. $C_(3)H_(6)$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$C_(3)H_(6)=3(CH_(2))$

D. $P_(4)O_(10)$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$P_(4)O_(10)=2(P_(2)O_(5))$

Hence, the empirical formula is the optionB. $NH_(3)$

Have a nice day!

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Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl5(g) → PCl3(g) + Cl2(g) P4(s) + 6Cl2(g) → 4PCl3(g) ΔH = -2439 kJ
4PCl5(g) → P4(s) + 10Cl 2(g) ΔH = 3438 kJ

Answers

The value of ΔH will be 249.75 KJ.

To obtain the reaction :

PCl5(g) → PCl3(g) + Cl2(g)

We have to add the given reactions and then divide it by 4.

(i) adding the reactions

P4(s) + 6Cl2(g) → 4PCl3(g) ΔH = -2439 kJ

4PCl5(g) → P4(s) + 10Cl 2(g) ΔH = 3438 kJ

we get:

4PCl5(g) → 4PCl3(g) + 4Cl2(g), ΔH = 3438 - 2439

(ii) dviding by 4

PCl5(g) → PCl3(g) + Cl2(g) , ΔH = (3438 - 2439)/4

ΔH = 249.75 kJ is the required enthapy.

Learn more about enthalpy:

brainly.com/question/12676996

Answer:

The value of $\Delta H$ for the desired reaction will be 249.75 KJ.

Explanation:

The desired reaction is shown below

$\textrm{PCl}_(5)\left ( g \right )\rightarrow \textrm{PCl}_(3)\left ( g \right )+\textrm{Cl}_(2)\left ( g \right )$

The desired reaction can be obtained by adding the given reactions and then dividing both sides by 4.

$P_(4)\left ( s \right )+6Cl_(2)\left ( g \right )\rightarrow 4PCl_(3)\left ( g \right ) \n4PCl_(5)\left ( g \right )\rightarrow P_(4)\left ( s \right )+10Cl_(2)\left ( g \right )$