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3 years ago

Water boils at 100Â°C at sea level. If the water in this experiment did not boil at 100Â°C, what could be the reason?

Chemistry

2 answers:

Alexeev081 [22]3 years ago

8 0

Answer:

You are not at sea level.

Explanation:

Assuming empirical observation shows boiling point 'below' 100°C then one concludes boiling experiment is being conducted above sea level where air pressure is lower than sea level air pressure. Generally, it is accepted in most literature publications that atmospheric pressure decreases ~1-inch per 1000-ft increase in elevation.

By definition, the boiling point of a liquid occurs when the atmospheric pressure equals fluid vapor pressure. Thus, if experimentally a water sample boils less than 100°C then one can only conclude the atmospheric pressure is lower than that needed to boil water at sea level atmospheric pressure; i.e., observed boiling point is at elevation greater than sea level.

One can calculate the boiling point of a liquid relative to sea level boiling point by using the Clausis-Clapeyron Equation.

ln(VP₂/VP₁) = ΔH(vap)/R(T₁ - T₂/T₁·T₂) solve for T₂ given Heat of Vaporization of water (ΔH(vap) = 28.2 Kj/mol) at specified temperature (T₁ at sea level = 273 Kelvin - standard) and R = 0.008314 Kj/mol-K).

slavikrds [6]3 years ago

3 0

The water isn’t at sea level

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from Avogadro's law

39g of K contains 6.02x10^23 atoms

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39=6.02x19^23

X=5.11×10^22

making X the subject of the formula

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2 years ago

How many molecules of zinc oxide are there in a 2 kg sample?

Novosadov [1.4K]

There are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample. Details about number of molecules can be found below.

<h3>How to calculate number of molecules?</h3>

The number of molecules of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

According to this question, there are 2000g of ZnO in a sample. Zinc oxide has a molar mass of 81.38 g/mol.

no of moles = 2000g ÷ 81.38g/mol

no of moles = 24.57mol

number of molecules = 24.57 × 6.02 × 10²³

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2 years ago

Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and

tatyana61 [14]

<u>Answer:</u> The equilibrium concentration of $PCl_5$ is 1.285 M.

<u>Explanation:</u>

The chemical equation for the decomposition of phosphorus pentachloride follows:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

The expression for equilibrium constant is given as:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

We are given:

$K_c=0.042$

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The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

$0.042=\frac{0.18\times 0.30}{[PCl_5]}$

$[PCl_5]=1.285$

Hence, the equilibrium concentration of $PCl_5$ is 1.285 M.

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