Answer:
According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of 
and keep same equilibrium constant
Explanation:
In this buffer following equilibrium exists -
So, is involved in the above equilibrium.
When a strong base is added to this buffer, then concentration of increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of and keep same equilibrium constant.
Therefore excess amount of combines with 
to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.
1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:
n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol
2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation
[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg
3) Transform mol to mass using the atomic mass:
10.4 mol Mg * 24.3 g/mol = 252.7 g of Mg.
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Co2
Explanation:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)
Answer:
Explanation:
The question is not complete, the cmplete question is:
Identify one type of noncovalent bond present in each solid.
1) Table salt (NaCl) 2) Graphite (repeating)
a. hydrogen bonds
b. ionic interactions
c. van der Waals interactions
d. hydrophobic interactions
Answer:
1) Table salt
b. ionic interactions
Ionic bond are formed between atoms with incomplete outermost shell. Some atoms add electrons to their outermost shell to make the shell complete hence making it a negative ion while some atoms loses their electron to make the outermost shell complete becoming a positive ion. In NaCl, sodium (Na) has 1 electron in its outermost shell which it transfers to Cl which has 7 electrons in the outermost shell. Hence after the bonding the outermost shell of the atoms become complete.
2) Graphite
c. Van Der Waals interaction
Van der waal forces are weak interaction between molecules that exist between close atoms. Carbon atoms in graphite planes have covalent bond, these graphite planes are known as graphenes. Bonds between graphenes are very weak and are van der waals forces.
