Answer:
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Explanation:
ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.
In our case we have:
mass of MgBr₂ = 12.41 g
volume of water (which is equal to the final solution volume) = 2.55 L
Now we devise the following reasoning:
if 12.41 g of MgBr₂ are dissolved in 2.55 L of water
then X g of MgBr₂ are dissolved in 1 L of water
X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂
if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻
then in 4.867 g of MgBr₂ we have Y g of Br⁻
Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)
4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
J.J. Thomson hypothesized and discovered that the atom was not the smallest unit of matter but that instead there were much smaller units. He discovered "sub-atomic particles" which make up atoms. The sub-atomic particle that Thomson discovered was the electron. He discovered this through a process of experiments testing cathode rays.
Answer:
1.
2.
Explanation:
1.Momentum is given as the product of mass by velocity of an object.
Momentum,
m=1,500kh, v=6m/s
2.Momentum,
m=7800kg, v=30m/s
new mass=7800+800=8600
As mass is increased, so does the resultant velocity as mass is directly proportional to velocity.
Answer: HCI + KOH → KCI + H20
Explanation:
HCI(aq) + KOH(aq) → KCI(aq) + H20(l)
Acid + base → Salt + Water.
The above is a neutralization reaction in which an acid, aqeous HCl reacts completely with an appropriate amount of a base, aqueous KOH to produce salt, aqueous KCl and water, liquid H2O only.
This is a neutralization reaction since, the hydrogen ion, H+, from the HCl is neutralized by the hydroxide ion, OH-, from the KOH to form the water molecule, H2O and salt, KCl only.
The volume that will occupy at STP is calculated as follows
by use of ideal gas equation
that is PV=nRT where n is number of moles calculate number of moles
n= PV/RT
p=0.75 atm
V=6.0 L
R = 0.0821 L.atm/k.mol
T= 35 +273= 308k
n=?
n= (o.75 atm x 6.0 L)/( 0.0821 L.atm/k.mol x 308 k)= 0.178 moles
Agt STP 1 mole= 22.4 L what obout 0.178 moles
= 22.4 x0.178moles/ 1moles =3.98 L( answer C)
