Answer:
Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.
Answer:
true
Explanation:
H2O : 2 covalent bonds are established between O atom and 2 H atom (H--O--H) by mutual shearing of their valence electrons.
<h3>Answer:</h3>
Strontium (Sr)
<h3>Explanation:</h3>
The condition given in statement is the presence of two valence electron. Hence, first we found the electronic configuration of given atoms as follow;
Rubidium [Kr] 5s¹
Strontium [Kr] 5s²
Zirconium [Kr] 4d² 5s²
Silver [Kr] 4d¹⁰ 5s¹
From above configurations it is cleared that only Strontium and Zirconium has two electrons in its valence shell.
We also know that s-block elements are more reactive than transition elements due to less shielding effect in transition elements hence, making it difficult for transition metals to loose electrons as compared to s-block elements. Therefore, we can conclude that Strontium present in s-block with two valence electrons is the correct answer.
Because a copper ion looses electrons, meaning it's negatively charged, and positives and negatives attract.
Answer:
<h2>D) 6</h2>
Explanation:
since, n = molar mass / empirical formula mass
Empirical formula mass = Total mass of atoms present in empirical formula
CHCl = 12+1+35.5
= 48.5
Given, Molar mass = 290.8 g.
So, n = 290.8/48.5
= 5.995 , that is approx 6.
So, Molecular formula = n × Empirical formula
= 6 × CHCl
= 
So, Number of C = 6
