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PilotLPTM [1.2K]
4 years ago
11

A container holds 20.0 grams of neon gas. Under the same conditions, how many grams of xenon would the container hold?

Chemistry
2 answers:
lara31 [8.8K]4 years ago
5 0
<span>A container that holds 20.0 grams of neon gas under the same conditions how many grams of xenon would the container hold? A container that holds 20.0 ...</span><span>
</span>
KatRina [158]4 years ago
4 0

Answer: 131 grams

Explanation:-

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

\text{Number of moles of neon}=\frac{20g}{20g/mol}=1mole

Thus 1 mole of Xenon will occupy the same volume as 1 mole of Neon.

Mass of xenon=moles\times {\text {Molar mass}}=1\times 131g/mol=131g

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What happens when the concentration of water inside a cell is lower than the concentration of water outside the cell?
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Answer:

Water outside the cell will flow inwards by osmosis to attain equilibrium

Explanation:

In the hypotonic environment, the concentration of water is greater outside the cell and the concentration of solute is higher inside. A solution outside of a cell has a lower concentration of solutes relative to the cytosol.

If concentrations of dissolved solutes are greater inside the cell, the concentration of water inside the cell is correspondingly lower. As a result, water outside the cell will flow inwards by osmosis to attain equilibrium.

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3 years ago
Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature
weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

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\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

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