Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]
Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess.
Follow the steps as
1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.
n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-
Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess
n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.
Total volume= V acid + V base= 28 ml + 60 ml = 98 ml
Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M
The answer is 0.009 M.
Answer:
Mass of barium sulfate = 8.17 g
Explanation:
Given data:
Mass of sodium sulfate = 4.98 g
Mass of barium sulfate produced = ?
Solution:
Na₂SO₄ + Ba(NO₃)₂ → BaSO₄ + 2NaNO₃
Moles of sodium sulfate:
Number of moles = mass/molar mass
Number of moles =4.98 g / 142.04 g/mol
Number of moles = 0.035 mol
Now we will compare the moles pf sodium sulfate and with barium sulfate.
Na₂SO₄ : BaSO₄
1 : 1
0.035 : 0.035
Mass of barium sulfate:
Mass = number of moles × molar mass
Mass = 0.035 mol ×233.4 g/mol
Mass = 8.17 g
Answer:
Tetrahedral
Explanation:
For the repulsion of the free electron pair theory, the shape of a molecule will be to repel the bonds and the free electrons on the central atom. In a molecule of carbon tetrachloride, the central atom (C) has no free electrons, so, the shape that repels better the charge is tetrahedral, as shown below.
The answer is B.
You can rule C out because divergent means moving away. Rule out A because there is an oceanic and continental plate, not 2 of the same type. Rule D out for the same reason.
