Arsenic, I believe. Metalloids fall in between metals and nonmetals (usually on the bold line separating the two on the periodic table). And since the metalloid in question has four electron shells and five valence electrons in the outermost shell, you can see that this element is arsenic
Hi. I can't help you if you don't have an equation. Please provide one.
95.611 g/mol that's the answers
Answer:
- The answer is the concentration of an NaOH = 1.6 M
Explanation:
The most common way to solve this kind of problem is to use the formula
In your problem,
For NaOH
C₁ =?? v₁= 78.0 mL = 0.078 L
For H₂SO₄
C₁ =1.25 M v₁= 50.0 mL = 0.05 L
but you must note that for the reaction of NaOH with H₂SO₄
2 mol of NaOH raect with 1 mol H₂SO₄
So, by applying in above formula
- (C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)
- C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M
<u>So, the answer is the concentration of an NaOH = 1.6 M</u>
