342.15 g/mol is the molar mass of Al2(SO4)3 Aluminium sulfate, This is what I found I hope this is right. Hope this helps;)
Explanation:
Elements in the same group have same number of valence electrons. And we know, the elements which have same number of valence electrons, have similar physical and chemical properties. Hence, the elements in the same group have similar physical and chemical properties.
Networks of feeding relationships is correct
Answer:
V₂ ≈416.7 mL
Explanation:
This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.
where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.
The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,
- V₁= 600 mL
- T₁= 360 K
- T₂= 250 K
Substitute the values into the formula.
- 600 mL /360 K = V₂ / 250 K
Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.
- 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
- 250 K * 600 mL/360 K = V₂
The units of Kelvin cancel, so we are left with the units of mL.
- 250 * 600 mL/360=V₂
- 416.666666667 mL= V₂
Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.
The volume of the balloon at 250 K is approximately 416.7 milliliters.
<u><em>The process of how we would obtain </em></u><u><em>ethanal</em></u><u><em> </em></u><u><em>free</em></u><u><em> from ethanol is described in the explanations below. </em></u>
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- In Chemistry, Ethanol undergoes oxidation in the presence of sodium dichromate plus sulphuric acid to yield ethanal and water.
The procedure for achieving this in the laboratory is as follows;
- Step 1; Measure a quantity of a solution of sodium dichromate acidified in a dilute sulphuric acid and pour into a test tube.
- Step 2; Add excess <em>ethanol</em>. This is because if we don't do so there will be plenty of oxidizing agent to carry out a second operation which changes the aldehyde to ethanoic acid. However, we need only the aldehyde.
- Step 3; When the aldehyde ethanal begins to form which will be evident by the change in the colour of solution from <em>orange to green</em>, then the mixture should be distilled from the test tube and tbethe aldehyde collevted so that it doesn't undergo additional oxidation into ethanoic acid.
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