Answer:
New temperature t2 = [1.28T− 273.15]° C
Explanation:
Given:
Volume v1 = 25 gram
New volume v2 = 25 + 7 = 32 gram
Constant pressure = p
Temperature t1 = T
Find:
New temperature t2
Computation:
Pv1/t1 = Pv2/t2
25 / T = 32 / t2
t2 = 1.28T
New temperature t2 = [1.28T− 273.15]° C
A = piR^2
a = pi × (8÷2)^2
a = pi × 4^2
a = pi × 16
a =50.27ft^2
Answer:
oh it's easy
Explanation:
Take the hydrate
N
a
2
S
2
O
3
∙
5
H
2
O
. Are there ionic forces between the
N
a
+
and the
S
2
O
2
−
3
and ion-dipole forces between the cation/anions and the water?
