Profundal
profundal zone is the coldest and deepest zone
Explanation:
Formula according to the radius ratio rule is as follows.

= 0.397
According to the radius ratio rule, as the calculated value is 0.397 and it lies in between 0.225 to 0.414. Therefore, it means that the type of void is tetrahedral.
Thus, we can conclude that the given compound is most likely to adopt closest-packed array with lithium ions occupying tetrahedral holes.
Assuming the balloon initially has volume of 0 when deflated, the total P(deltaV) = (1.00 atm)(550,000 ft^3) = 550,000 atm-ft^3. To convert into work units, we can first convert ft^3 to L:
(550,000 atm-ft^3)(1 m/3.28 ft)^3
= (15,586.2 atm-m^3)
Then we convert to L:
(15,586.2 atm-m^3)(1000 L/m^3)
= 15,586,200 atm-L
Then we convert to J:
(15,586,200 atm-L)(101.325 J / 1 atm-L)
= 1.579 x 10^9 J
Answer:
0.036 M
Explanation:
To do this, let's mark the dye as D and bleach as B.
We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.
The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.
To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:
moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles
moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles
Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:
moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles
These are the moles presents of B after the reaction has been made. The concentration of the same will be:
[B] = 3.597x10⁻⁴ / 0.010
[B] = 0.0357
With 2 SF it would be:
[B] = 3.6x10⁻² M
Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


Now to calculate the molarity of original solution:


Thus the molar concentration of sulfuric acid in the original sample is 1.943 M