Answer:
a beam of electrons emitted from the cathode of a high-vacuum tube.
Explanation:
Explanation:
According to Buoyance equation,
m = ![[m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]](https://tex.z-dn.net/?f=%5Bm%27%20%5Ctimes%20%5Cfrac%7B1%20-%20%5Cfrac%7Bd_%7Ba%7D%7D%7Bd_%7Bw%7D%7D%7D%7B1%20-%20%5Cfrac%7Bd_%7Ba%7D%7D%7Bd%7D%7D%5D)
where, m = true mass
m' = mass read from the balance = 17.320 g
= density of air = 0.0012 g/ml
= density of the balance = 7.5 g/ml
d = density of liquid octane = 0.7025 g/ml
Now, putting all the given values into the above formula and calculate the true mass as follows.
m =
= ![[17.320 g \times \frac{1 - \frac{0.0012 g/ml}{7.5 g/ml}}{1 - \frac{0.0012 g/ml}{0.7025}}]](https://tex.z-dn.net/?f=%5B17.320%20g%20%5Ctimes%20%5Cfrac%7B1%20-%20%5Cfrac%7B0.0012%20g%2Fml%7D%7B7.5%20g%2Fml%7D%7D%7B1%20-%20%5Cfrac%7B0.0012%20g%2Fml%7D%7B0.7025%7D%7D%5D)
= 
= 17.317 g
Thus, we can conclude that the true mass of octane is 17.317 g.
Answer:
Explanation:
mass of one virus = 9.0 x 10⁻¹² mg
mass of one mole = 6.02 x 10²³ x mass of one virus
= 6.02 x 10²³ x 9.0 x 10⁻¹²
= 54.18 x 10¹¹ mg
= 54 x 10⁸ g .
= 54 x 10⁵ kg .
b )
let n be no of moles of virus that will be equal to weight of oil tanker
n x 54 x 10⁵ = 3 x 10⁷
n = 5.5555
rounding off to 2 significant figure
5.6 moles Ans .
Molar mass KCl = <span>74.5513 g/mol
Number of moles:
21.9 / 74.5513 => 0.293 moles
Volume = 869 mL / 1000 => 0.869 L
Molarity = moles / Volume
Molarity = 0.293 / 0.869
=> 0.337 M</span>
Answer:
ionic, they would make an ion
Explanation:
