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3 years ago

An unknown gas shows a density of 3 g per litre at

Chemistry

1 answer:

Jlenok [28]3 years ago

5 0

Answer:

Mass of 1 litre of this gas at 273 degree Celsius and 1140 mm Hg pressure is

3 grams. So, (T1) = (273 + 273) degree Absolute = 546 degree Absolute and

(P1) = (760 + 1140) mm Hg = 1900 mm Hg and (V1) = 1 Litre

First, we have to find out how much volume does it occupy at NTP; (T2) = 273 degree Absolute and (P2) = 760 mm Hg.

So, (V2) = (1900*1 / 546)*(273 / 760) = 1.25 litre.

So, the mass of 1.25 litre volume of this gas = 3 grams

Therefore, the mass of 22.4 litre volume of this gas = (3*22.4 / 1.25) grams

= 53.76 grams.

So, the gram molecular mass of the gas is 53.76.

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6 0

4 years ago

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The measure of distance from one point to another is?

olasank [31]

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3 years ago

Consider the reaction:

stich3 [128]

Answer:

$\large \boxed{\text{-851.4 kJ/mol}}$

Explanation:

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}$

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4 years ago

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JulsSmile [24]

Answer:

Identify the Problem

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3 years ago

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What is the limiting reactant when 5.6 moles of aluminum react with 6.2 moles of water?

Arlecino [84]

here we have to choose the limiting reactant in the given unbalanced reaction.

The 5.6 moles of aluminum will be the limiting reactant in this reaction.

The reagent or reactant which is totally consume in a reaction is called the limiting reagent or reactant.

The unbalanced reaction is Al + H₂O → Al₂O₃ + H₂

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As we can see that 2 moles of aluminium (Al) reacts with 3 moles of water (H₂O). Thus 5.6 moles of Al reacts with $\frac{2}{3}$ ×5.6 = 3.733 moles of water.

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3 years ago