<u>Answer:</u> The volume of barium chlorate is 195.65 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
Given mass of barium chlorate = 25.0 g
Molar mass of barium chlorate = 304.23 g/mol
Molarity of solution = 0.420 mol/L
Volume of solution = ?
Putting values in above equation, we get:
Hence, the volume of barium chlorate is 195.65 mL
Copper has the highest thermal conductivity as compared to other metals in the periodic table.
<h3>What is thermal conductivity?</h3>
Thermal conductivity refers to the rate at which heat is transferred by conduction through a unit cross-section area of a material. When a temperature increases at one end it moves to the other end through conduction.
So we can conclude that Copper has the highest thermal conductivity as compared to other metals in the periodic table.
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a) 
→ 
b) 
→ 
<h3>
What are half-reactions?</h3>
The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.
a)
→
= 
= -0.83 - (-2.71) =1.88V
b)
→
= 
=-0. - (0.8) =-0.8V
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Atmosphere is made of a mixture of gases. Each individual gas exerts a pressure known as the partial pressure.
The sum of the partial pressures exerted by each gas in the mixture is equal to the total pressure exerted by the mixture of gases as a whole.
The partial pressure depends on the composition of the gas in the mixture.
In this case the total pressure - 870 mmHg
21% of atmosphere is oxygen
partial pressure of Oxygen = 870 mmHg x 21% = 182.7 mmHg
Answer:
0.2042 M is the original concentration of 
(aq) in the titrating solution.
Explanation:
Mass of 
= 0.217 g
Moles of 
1 mole of have 2 mole of As and 1 mole of 
have 1 mole of As.
So, from 1 mole of we will have 2 moles of
Then from 0.001096 mol of :
of
According to reaction, 1 mole of reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of
of cerium (IV) ions.
Volume of the acidic cerium{IV) sulfate = 21.47 ml =0.02147 L
1 mL = 0.001 L
![[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M](https://tex.z-dn.net/?f=%5BCe%5E%7B4%2B%7D%5D%3D%5Cfrac%7B0.004384%20mol%7D%7B0.02147%20L%7D%3D0.2042%20M)
0.2042 M is the original concentration of (aq) in the titrating solution.
