Döbereiner grouped the known elements into <em>triads</em> (sets of three) so that
• The <em>atomic mass of the middle element</em> was approximately the average of the other two
• The <em>chemical properties of the middle element</em> were between those of the other two
• The <em>physical properties of the middle element</em> were between those of the other two
One example of a triad is Li – Na – K.
(a) Atomic mass of Na = 23.0 u
Average atomic mass of Li and K = (6.9 u + 39.1 u)/2 = 46.0 u/2 = 23.0 u
(b) Li reacts slowly with water. Na reacts rapidly. Potassium reacts violently.
(c) Melting point of Na = 371 °C.
Average melting point of Li and K = (454 °C + 330 °C)/2 = 784 °C/2
= 392 °C
Since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. The concentration can be measured in several units. Generally, concentration is expressed in molarity, molality, mass concentration units or percentage.
Now we are asked to find the amount concentration of calcium ions and acetate ions in a 0.80 mol/L solution of calcium acetate. The formula of calcium acetate is Ca(CH3COO)2.
Thus;
Ca(CH3COO)2(aq) ----> Ca^2+(aq) + 2CH3COO^-(aq)
It then follows that since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
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Answer:
Molecular solids and covalent network solids are two types of solid compounds. The key difference between molecular solid and covalent network solid is that <em>molecular solid forms due to the action of Van der Waal forces </em>where as <em>covalent network solid forms due to the action of covalent chemical bonds.</em>
hope this helps
Answer:
(A) N4H6 (B) H2O (C) LiH (D) C12H26
Explanation:
The given compounds have been arranged from left to right in order of increasing percentage by mass of hydrogen.
The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.
