Answer:
B
Explanation:
as it depends on electronic configuration
Answer:
197mL of 0,506M HCl
Explanation:
The reaction of HCl + BaCO₃ is:
BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.
The moles of BaCO₃ in 9,85 g are:
9,85 g of BaCO₃ × 
= <em>0,0499 moles of BaCO₃</em>
As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:
0,0499 moles of BaCO₃ × 
=<em> 0,0998 moles of HCl</em>
If you have a 0,506M HCl, you need to add:
0,0998 moles of HCl× 
= 0,197 L ≡ 197mL
I hope it helps!
Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.
Answer:
There are 6.022x10^23 molecules
Answer:
no
Explanation:
"Alkali metals are among the most reactive of all metals, which makes them suitable for specific and limited uses.
Alkali metals include lithium, sodium, potassium, rubidium, cesium and francium. These metals have large atomic radii and generally lose electrons during reactions. "
- Reference
