Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
Charles law gives the relationship between temperature and volume of gases. It states that the volume of gas is directly proportional to temperature at constant pressure.
V / T = k
where V - volume and T - temperature in Kelvin and k - constant

where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - 20 °C + 273 = 293 K
substituting these values in the equation

T = 586 K
temperature in celsius = 586 K - 273 = 313 °C
new temperature is 313 °C
Answer:
The 2 would be placed in front of the reactant Na and in front of the product NaCl
2Na + Cl2 = 2NaCl
Explanation:
This is because the ratio of elements needs to be balanced on both sides.
On the reactants side, there are 2 Na molecules and 2 Cl molecules
On the products side, there are 2 Na molecules and 2 Cl molecules
So, now the equation is balanced