Answer:
Group 1 ( The Leftmost)
More Information :
The alkali metals are six chemical elements in Group 1, the leftmost column in the periodic table.
They are :
- lithium (Li)
- sodium (Na)
- potassium (K)
- rubidium (Rb)
- cesium (Cs)
- francium (Fr).
Answer: An amount of
heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.
Explanation:
Given: mass of lead = 4.64 kg
Convert kg into grams as follows.



The standard value of specific heat of lead is
.
Formula used to calculate heat is as follows.

where,
q = heat energy
m = mass of substance
C = specific heat of substance
= change in temperature
Substitute the value into above formula as follows.

Thus, we can conclude that
heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.
Answer:
89.4%
Explanation:
Initially, there is 5.0 of the acetanilide in 100 mL of water, then the solution is chilled at 0ºC. The solubility represents the amount that the solvent (water) can dissolve of the solute (acetanilide). So, at 0ºC, 100 mL of water can dissolve till 0.53 g of the compound, the rest will precipitate and will be recovered.
So, the mass that is recovered is 5.0 - 0.53 = 4.47 g
The percent recovery is:
(4.47/5)x100% = 89.4%
If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.
Question options:
a) 2.05
b) 0.963
c) 0.955
d) 1.00
Answer:
b) 0.963
Explanation:
H2SO4→ HSO4- + H3O+
HSO4- + H2O ⇌ SO42- + H3O+
Construct ICE table:
HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)
I 0.1 solid & 0 0.1
C -x liquid + x + x
E 0.1 - x are ignored x 0.1 + x
Calculate x
Ka = products/reactants
= ![\frac{[SO42-] [H3O+]}{[HSO4-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO42-%5D%20%5BH3O%2B%5D%7D%7B%5BHSO4-%5D%7D)
0.011 = 
0.011 x (0.1 -x) = o.1x + x^2
0.0011 - 0.011 x - o.1x - x^2 = 0
0.0011 - 0.011 x - x^2 = 0
Use formula to solve for quadratic equation
x =
/ 2a
a = -1, b = -0.111, c = 0.001
Solve for x
x =
/ 2(-1)
x = 0.111 +,-
/ -2
x = 0.111 +,-
/ -2
x = 
x =
, x = 
x =
, x = 
x = - 0.12015 , x = 0.00915
x cannot be negative, so
x = 0.00915 M
Calculate [H3O+]
[H3O+] = 0.1 M + x
[H3O+] = 0.1 M + 0.00915 M
[H3O+] = 0.10915 M
Clculate pH
pH = - log [ H3O+]
pH = - log [ 0.10915]
pH = 0.963