Answer:
cesium
In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). In fact, for the alkali metals (the elements in Group 1), the ease of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron
Explanation:
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J
(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal
(rounded to three significant figures)
95.6 cal
are needed.
the answer is A. SnCl2 . 2H2O
Answer: The balanced equation is 
.
Explanation:
The given reaction equation is as follows.
Number of atoms present on reactant side are as follows.
- Li = 1
- H = 1
= 1
Number of atoms present on product side are as follows.
- Li = 1
- H = 2
- = 1
To balance this equation, multiply Li by 2 and 
by 2 on reactant side. Also, multiply 
by 2 on product side.
Hence, the equation can be rewritten as follows.
Now, number of atoms present on reactant side are as follows.
- Li = 2
- H = 2
- = 2
Number of atoms present on product side are as follows.
- Li = 2
- H = 2
- = 2
As there are same number of atoms on both reactant and product side. Hence, the equation is now balanced.
Thus, we can conclude that the balanced equation is .
A hypothesis is how you think the experiment is going to end
