Answer:
58.44 g/mol The Molarity of this concentration is 0.154 molar
Explanation:
the molar mass of NaCl is 58.44 g/mol,
0.9 % is the same thing as 0.9g of NaCl , so this means that 100 ml's of physiological saline contains 0.9 g of NaCl. One liter of physiological saline must contain 9 g of NaCl. We can determine the molarity of a physiological saline solution by dividing 9 g by 58 g... since we have 9 g of NaCl in a liter of physiological saline, but we have 58 grams of NaCl in a mole of NaCl. When we divide 9 g by 58 g, we find that physiological saline contains 0.154 moles of NaCl per liter. That means that physiological saline (0.9% NaCl) has a molarity of 0.154 molar. We can either express this as 0.154 M or 154 millimolar (154 mM).
D make your car easy to spot in daytime because they do not illuminate a lot
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer:
t = 25.10 sec
Explanation:
we know that Avrami equation
here Y is percentage of completion of reaction = 50%
t is duration of reaction = 146 sec
so,
taking natural log on both side
ln(0.5) = -k(306.6)
for 86 % completion
t = 25.10 sec
