A scientist needs to check several parameters before coming
to a conclusion about the amount of water pollution. The scientists needs to
check the amount of dissolved oxygen in the water, temperature of the water,
the clarity of the water, the PH level of the water and also the amount of
bacteria present in the water. There may be other criteria’s, but the mentioned
ones are enough to gauge the amount of pollution in the water. Scientists often
takes fish and aquatic plants from the water to be tested to check the amount
of pollution indirectly affecting these species.
Answer: T2= 962.2 K
Explanation:
The ideal gases is often written like PV=nRT, where P is pressure, V is volume, n is moles, R is the universal constant of the gases and T is Temperature.
So, in this problem there is a container that is a closed system, therefore n is constant and volume too. The initial point is 1 and the final point is 2, so
V1=V2 ⇒

Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Solubility of a compound in water can be referred to as the amount of the compound that can be dissolved in 1 L of the solvent (water) at any given temperature. Solubility of a compound can be expressed in the units of g/L or mg/L.
Given that the solubility of calcium carbonate in water = 14 mg/L
We have to calculate the volume of water that can dissolve 11 g of calcium carbonate.
Converting 11 g calcium carbonate to mg:

Volume of water that would dissolve 11000 mg calcium carbonate
= 
=785.7 L
Rounding the volume 785.7 L to two significant figures, we get 790 L water.
Therefore, we would need 790 L water to completely dissolve 11 g of calcium carbonate.