Answer: 
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte like urea)
= freezing point constant = 
m= molality
Weight of solvent (X)= 950 g = 0.95 kg
Molar mass of non electrolyte (urea) = 60.06 g/mol
Mass of non electrolyte (urea) added = ?
Thus urea was dissolved.
The SI unit for the amount of substance present is the mole.
The mole is defined as the amount of substance that has the same amount of particles as there are atoms in 12 grams of carbon-12. Mathematically, the moles of a substance may be computed using:
moles present = mass of substance / molecular mass of substance
Answer:
Column 17 halogens
Explanation:
To answer the question we need to know the following;
What are valence electrons?
- Valence electrons are outermost electrons in the outermost energy level of an atom of any element.
What are halogens?
- Halogens are group 7 elements that have seven valence electrons. They include, chlorine, fluorine, iodine, etc
What is the reactivity of halogens?
- Halogens are the most reactive non-metals in the periodic table. They react by gaining one electron to attain a stable configuration.
Answer:
An atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)
Explanation:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other.
For example,
Al atom has 13 protons and 13 electrons. The number of positive and negative charge is equal thus it will be neutral atom.
While the atom of Al which have 13 proton and 10 electron is not neutral. The positive charge is greater than negative by 3. Which means 3 electrons are lose by Al atom and form cation "Al⁺³".
Thus an atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)
There will be 7.5 g of Be-11 remaining after 28 s.
If 14 s = 1 half-life, 28 s = 2 half-lives.
After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.
After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.
In symbols,
<em>N</em> = <em>N</em>₀(½)^<em>n</em>
where
<em>n</em> = the number of half-lives
<em>N</em>₀ = the original amount
<em>N</em> = the amount remaining after <em>n</em> half-lives
