Answer:
The more acidic the solution the faster it rusts. More Na = more rust
Explanation:
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Answer: (a) The products are
and AgCl. The balanced equation is
.
(b) The type of reaction is double displacement reaction.
Explanation:
(a) The complete reaction equation will be as follows.
![BeCl_{2}(aq) + 2AgNO_{3}(aq) \rightarrow Be(NO_{3})_{2}(aq) + 2AgCl(s)](https://tex.z-dn.net/?f=BeCl_%7B2%7D%28aq%29%20%2B%202AgNO_%7B3%7D%28aq%29%20%5Crightarrow%20Be%28NO_%7B3%7D%29_%7B2%7D%28aq%29%20%2B%202AgCl%28s%29)
Number of atoms on reactant side are as follows.
- Be = 1
- Cl = 2
- Ag = 2
= 2
Number of atoms present on product side are as follows.
- Be = 1
- Cl = 2
- Ag = 2
= 2
Here, number of atoms on both reactant and product side are the same. Hence, it is a balanced chemical equation.
(b) A reaction in which two reactant species tend to change their ions with each other and form new compounds. This type of reaction is called double displacement reaction.
As the given reaction is as follows.
![BeCl_{2}(aq) + 2AgNO_{3}(aq) \rightarrow Be(NO_{3})_{2}(aq) + 2AgCl(s)](https://tex.z-dn.net/?f=BeCl_%7B2%7D%28aq%29%20%2B%202AgNO_%7B3%7D%28aq%29%20%5Crightarrow%20Be%28NO_%7B3%7D%29_%7B2%7D%28aq%29%20%2B%202AgCl%28s%29)
This shows that ions are being exchanges by the reactants
and
. The ions are
,
,
and
.
Therefore, this reaction is a double displacement reaction.
True!! they can be measured without changing chemical identity.
Answer:
Temperature of freezing for the solution is. 3.3°C
Explanation:
This is an excersise of coligative property (freezing low depression)
ΔT = Kf . m
Mass of solute (biphenyl) → 0.617 g
Mass of solvent (cycloexane) → 25 g
m = molality (mol of solute in 1kg of solvent)
Let's find out moles of solute:
Moles = Mass / Molar mass
Moles = 0.617 g / 154 g/m → 0.004 moles
To know molalilty, let's convert 25 g to kg
25 g = 0.025 kg
moles / kg = molality
0.004 m / 0.025 kg = 0.16m
ΔT = T°freezing pure solvent - T° freezing solution
6.5°C - T° freezing solution = 20°C/m . 0.16m
6.5°C - T° freezing solution = 3.20°C
6.5°C - 3.20°C = T° freezing solution → °3.3 °C