<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
------------------------------
1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>

7 0

2 years ago

Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.

Ket [755]

Answer:

c. 2,2-dichloropentane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:

$CH\equiv C-CH_2-CH_2-CH_2$

Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:

$CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2$