Answer:
On treatement of 2 -with hydrogen gas and lindlar catalyst, the major product obtained is cis-2butene.
Ice and Vaseline. Those are two melting compounds. When melting ice it turns into water and when melting Vaseline it melts into a oily liquid form. You can also freeze both of them and you can melt them again.
Answer:
The answer to your question is Argon
Explanation:
Electron configuration given 1s² 2s² 2p⁶ 3s² 3p⁶
To find the element whose electron configuration is given, we can do it by two methods.
Number 1. Sum all the exponents the result will give you the atomic number of the element.
2 + 2 + 6 + 2 + 6 = 18
The element with an atomic number of 18 is Argon.
Number 2. Look at the last terms of the electronic configuration
3s² 3p⁶
Number three indicates that this element is in the third period in the periodic table.
Sum the exponents 2 + 6 = 8
Number 8 indicates that this element is the number 8 of that period without considering the transition elements.
The element with these characteristics is Argon.
Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M