Answer:
Be,Mg,Ra etc
Explanation:
It should be palced in group 2A because as it reacts with chlorine in ratio of 1:2 . It's valancy is 2 and is metal as it react with non metal donating two electrons .
one more thing it fits there orderly
Answer:
Boron has a larger radius and the protons in carbon exert more pull.
Explanation:
Remember than elements have greater radius as they are closer to the bottom left corner, so boron would have the larger radius here. Carbon has a smaller radius, which makes it easier for the protons in carbon to exert more pull.
Answer:
A: 1,2-dimethylcyclopropane
Explanation:
The possible cyclic structure with formula C₅H₁₀ are shown in the image.
A is a cyclic compound. On monochlorination, A yields 3 products.
To have 3 products on monochlorination, there should be three different carbon atoms.
Considering structure 1, all carbons have same nature, thus only one product will be formed and thus not a structure of A.
Considering structure 2, there are two different carbon atoms, thus two different structure are formed and thus not a structure of A.
Considering structures 3 and 4 , there are four different carbon atoms, thus four products will be formed and either of them are not a structure of A.
Considering structure 5, there are three different carbon atoms, thus three different structure are formed and thus the A is structure 5.
Answer:
0.1082M of Barium Hydroxide
Explanation:
KHP reacts with Ba(OH)2 as follows:
2KHP + Ba(OH)2 → 2H2O + Ba²⁺ + 2K⁺ + 2P²⁻
<em>Where 2 moles of KHP reacts per mole of barium hydroxide</em>
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To solve this question we must find the moles of KHP in 1.37g. With these moles and the reaction we can find the moles of Ba(OH)2 and its molarity using the volume of the solution (31.0mL = 0.0310L) as follows:
<em>Moles KHP -Molar mass: 204.22g/mol-</em>
1.37g * (1mol / 204.22g) = 0.006708 moles KHP
<em>Moles Ba(OH)2:</em>
0.006708 moles KHP * (1mol Ba(OH)2 / 2mol KHP) =
0.003354 moles Ba(OH)2
<em>Molarity:</em>
0.003354 moles Ba(OH)2 / 0.0310L =
<h3>0.1082M of Barium Hydroxide</h3>