Iconic is most soluble in water
Explanation:
When two or more substances are mixed together in such a way that their particles are evenly distributed throughout the mixture then it is known as a solution.
Basically, a solution is a homogeneous mixture which can be present in any phase.
A solution contains both solute and solvent.
A colloid is defined as a substance in which solute particles are microscopically distributed throughout the solvent. In a colloid both solvent and solute are present in liquid phase.
Hence, a solution can be easily distinguished from colloids.
Therefore, we can conclude that following statements about solutions are true:
- They contain solutes and solvents.
- Their particles must be evenly distributed.
- They may contain solid, liquid, and gas simultaneously.
- They are always easy to distinguish from colloids.
- They are homogeneous matter.
Answer:
9.7 g
Explanation:
From the question,
Note: The molar volume of all gas at stp is 22.4 dm³ or 22.4 L
1 mol of oxygen gas (O₂) at stp = 22.4 dm³
X mole of oxygen gas (O₂) at stp = 6.8 L
X = (1 mol×6.8 L)/22.4 L
X = 0.3036 mol.
But,
Number of mole (n) = mass (m)/molar mass (m')
n = m/m'
m = n×m'.................. Equation 2
Where n = 0.3036 mol, m' = 32 g/mol
Substitute into equation 2
m = 0.3036×32
m = 9.7 g
Answer:
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).
Explanation:
The reaction between an acid and a base is called neutralization, forming a salt and water.
Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.
When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:
V acid *M acid = V base *M base
where V represents the volume of solution and M the molar concentration of said solution.
In this case:
- V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
- M acid= 0.129 M
- V base= ?
- M base= 0.135 M
Replacing:
0.0137 L* 0.129 M= V base* 0.135 M
Solving:

V base=0.0131 L = 13.1 mL
<u><em>
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>
Answer:

Explanation:
Given that:
The Half-life of
=
is less than that of 
Although we are not given any value about the present weight of
.
So, consider the present weight in the percentage of
to be y%
Then, the time elapsed to get the present weight of
= 
Therefore;

here;
= Number of radioactive atoms relating to the weight of y of 
Thus:

--- (1)
However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of
to be = 
Then:
---- (2)
here;
= Number of radioactive atoms of
relating to 3.0 a/o weight
Now, equating equation (1) and (2) together, we have:

replacing the half-life of
=
( since
)
∴

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o
Thus, The time elapsed is 