<h2>Hello!</h2>
The answer is:
The new volume is equal to 206.5 L.
<h2>Why?</h2>
To solve this problem, we need to assume that the pressure is constant, and use the Charle's Law equation, so, solving we have:

We are given:

Then, using the Charle's Law equation, we have:


Hence, we have that the new volume is equal to 206.5 L.
Have a nice day!
A. 0.0105 kg
b. 1570 m
c. 3.5x10^-6
d. 3500000
e. 1000 mL
f. 0.000358 m3
g. 548.6 cm3
Answer:
Neutralization
Explanation:
When an acid react with base it form the salt and water. The reaction is also called neutralization reaction because both neutralize each other.
In neutralization reaction equal amount of acid and base react to neutralize each other and equal amount of water and salt are formed. When pH does not reach to 7 its means there is less amount of one of reactant which is not fully neutralize.
Neutralization reactions are also used as first aid. For example when someone is dealing with HCl for cleaning purpose of toilet and get touched. It is advised to neutralize it with soap, milk or egg white.
Example:
Hydrochloric acid when react with the sodium hydroxide, a salt sodium chloride and water are formed.
Chemical equation:
HCl + NaOH → NaCl + H₂O
Titration:
Neutralization reactions are also used to determine the concentration of solution. Titration is a quantitative technique in which acid or base is gradually added into the solution whose concentration is to be determine until the neutral point is reached.
Answer:
C. Hair dryer
Explanation:
Alternating Current
The other kind of electricity is called alternating current, or AC. Power plants produce this
type of electricity. It is carried by large power lines from the plant to your home. When you
plug a TV, a lamp, or a hair dryer into the wall socket, you are using this type of electricity.
Answer: 1.
moles
2. 90 mg
Explanation:

According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus
moles of ozone is removed by =
moles of sodium iodide.
Thus
moles of sodium iodide are needed to remove
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide=
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of
.