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:
Across a period, effective nuclear charge increases as electron shielding remains constant. A higher effective nuclear charge causes greater attractions to the electrons, pulling the electron cloud closer to the nucleus which results in a smaller atomic radius. ... This results in a larger atomic radius.
Explanation:
Now lets d8
A.a low basket with plastic liner
Answer:
"Hydrogen is still available outside the core, so hydrogen fusion continues in a shell surrounding the core. The increasingly hot core also pushes the outer layers of the star outward, causing them to expand and cool, transforming the star into a red giant."
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g
