The volume of N₂ at STP=56 L
<h3>Further explanation</h3>
Given
2.5 moles of N₂
Required
The volume of the gas
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.
So for 2.5 moles gas :
Answer:
Purpose: To become familiar with the techniques for separation of amixture of solids.
Explanation:
a mixture of pure substances. If you have a mixture of tennis ballsand marbles (not pure substances by the way), it would be easy toseparate the mixture. However, it is more difficult to separate asand (also not a pure substance) and salt mixture. Even with verygood tweezers and a magnifying glass, it would be extremelytedious. You could take advantage of the fact that salt dissolvesin water and sand does not. To separate iron powder from an ironand sand mixture you can take advantage of the magnetic propertiesof iron and separate the mixture.
To summarize a complete procedure for separating a mixture ofseveral substances, it is best to prepare a flow chart. A flowchartis a schematic representation of an algorithm or a stepwiseprocess, showing the steps as boxes of various kinds, and theirorder by connecting these with arrows. Flowcharts are used indesigning or documenting a process.
Answer:
6 x 10⁶ g Fe
Explanation:
Step 1: Set up dimensional analysis
7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)
Step 2: Multiply, divide, and cancel out units
atoms Fe and atoms Fe cancel out.
mol Fe and mol Fe cancel out.
We should be left with g Fe.
7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe
116279(55.85) = 6.49 x 10⁶ g Fe
Step 3: Sig figs
There is only 1 sig fig in this problem.
6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe
Answer:
=154.8 J
Explanation:
The rise in temperature is contributed by the change in temperature.
Change in enthalpy = MC∅, where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.
Change in temperature = 100.0°C-20.0°C=80°C
ΔH=MC∅
The specific heat capacity of gold= 0.129 J/g°C
ΔH= 15.0g×0.129J/g°C×80°C
=154.8 J
