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2 years ago

The empirical formula is:

Chemistry

1 answer:

Hunter-Best [27]2 years ago

5 0

Answer:

C. the relative number of atoms of each element, using the lowest whole ratio.

Explanation:

The empirical formula is how we simplify the whole formula to simplify it to its smallest indivisible parts.

It is definitely not the actual number of atoms. If you see an empirical formula, don't think that it's the full thing.

It is also not a representation of a compound to show its atoms' arrangement: this would be a Lewis dot structure, or a ball and stick model, or something similar. We don't use the empirical formula for this purpose.

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How has СOVID-19 demonstrated the process of the scientific method for the public?

Vesnalui [34]

Answer:

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3 years ago

__C2H4+__O2-->__CO2+__H2O

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Answer:

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3 years ago

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

6 0

3 years ago

Consider the reaction.

Stella [2.4K]

Answer:

when the rates of the forward and reverse reactions are equal

Explanation:

In a chemical system, the reaction reaches a dynamic equilibrium when the rate of formation of product equals the rate of formation of reactants. This implies that both the forward and revered(backwards) reaction are occurring at the same rate.

5 0

3 years ago

A 5.000 g sample of Niso4 H2O decomposed to give 2.755 g of anhydrous NiSO4.

Vinvika [58]

Answer:

a) 7.0.

b) Nickel sulfate hepta hydrate.

c) 280.83 g/mol.

d) 44.9%.

Explanation:

a) What is the formula of the hydrate?

The mass of the hydrated sample (NiSO₄.xH₂O) = 5.0 g,

The mass of the anhydrous salt (NiSO₄) = 2.755 g,

The mass of water = 5.0 g - 2.755 g = 2.245 g.

∴ no. of moles of water = mass/molar mass = (2.245 g)/(18.0 g/mol) = 0.1247 mol.

∴ no. of moles of anhydrous salt (NiSO₄) = mass/molar mass = (2.755 g)/(154.75 g/mol) = 0.0178 mol.

∴ water of crystallization in the sample (x) = no. of moles of water/no. of moles of anhydrous salt (NiSO₄) = (0.1247 mol)/(0.0178 mol) = 7.0.

b) What is the full chemical name for the hydrate?

The name of the salt (NiSO₄.7H₂O) is Nickel sulfate hepta hydrate.

c) What is the molar mass of the hydrate?

(NiSO₄.7H₂O)

The molar mass = molar mass of NiSO₄ + 7(molar mass of H₂O) = (154.75 g/mol) + 7(18.0 g/mol) = 280.83 g/mol.

d) What is the mass % of water in the hydrate?

The mass % of water = (mass of water)/(mass of hydrated sample) x 100 = (2.245 g)/(5.0 g) x 100 = 44.9%.

8 0

3 years ago