Your answer to this question would be B. packaging waste.
hope this helps!
This is an exception to the general electronegativity trend. It can be explained by looking at the electron configurations of both elements.
<span>Be:[He]2<span>s2
</span></span><span>B:[He]2<span>s2</span>2<span>p1
</span></span>
When you remove an electron from beryllium, you are taking away an electron from the 2s orbital. When you remove an electron from boron, you are taking an electron from the 2p orbital. The 2p electrons have more energy than the 2s, so it is easier to remove them as they can more strongly resist the effective nuclear charge of the nucleus.
Answer:
16
Explanation:
Group two elements are alkaline earth metal.
All these have two valance electrons. In order to achieve noble gas configuration it loses its two valance and get complete octet.
Reaction with group 16.
Reaction with oxygen,
They react with oxygen and form oxide.
2Ba + O₂ → 2BaO
2Mg + O₂ → 2MgO
2Ca + O₂ → 2CaO
this oxide form hydroxide when react with water,
BaO + H₂O → Ba(OH)₂
MgO + H₂O → Mg(OH)₂
CaO + H₂O → Ca(OH)₂
With sulfur,
Mg + S → MgS
Ca + S → CaS
Ba + S → BaS
Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
Answer:
The answer is Near The Ocean Ridges.
