Answer:
decrease the volume of the cylinder.
Explanation:
In order to be able to solve this question we have to understand what Boyle's law is. According to Boyle's law; at constant temperature the pressure of a given mass of gas is inversely proportional to to its volume.
The Boyle's law shows us the relationship between the pressure and the volume. So, the Important thing to note hear is that if the volume in a container is decreased then the pressure will increase (and vice versa) due to the fact that as the volume decreases the particles in that container makes more collision which will make the pressure to increase.
Since, the piston is moveable it means that we can decrease and increase the volume in the cylinder. So, if the decrease the volume of the cylinder then we will have an increase in the pressure of the gas below the piston.
for it to be balanced in this case would be " <em>4</em> C6H6 + <em>6</em> CI2 = <em>3</em> C6H5CI + <em>9</em> HCI" therefore it's be a <u>Double Replacement</u>
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)
Answer:
0.14 moles of Fe₂O₃
Explanation:
Given parameters:
Number of moles of Fe = 0.27moles
Unknown:
Number of moles of Fe₂O₃ = ?
Solution:
To solve this problem, we are going to the work from the known specie to the unknown using their number of moles.
We first obtain the balanced equation of the reaction;
4Fe + 3O₂ → 2Fe₂O₃
The equation above is balanced;
4 moles of Fe produced 2 moles of Fe₂O₃
0.27 moles of Fe will produce 
= 0.14 moles of Fe₂O₃
Answer:
T2 = 2843.1 oK. This is a huge temperature. Check it for errors.
Explanation:
Remark
This is the same question as the other one I've answered. Only the numbers have been altered.
Givens
v1 = 56 mL
P1 = 1 atm
T1 = 273o K
v2 = 162
P2 = 3.6 atm
T2 = ?
Formula
Vi * P1 / T1 = V2 * P2/T2
Solution
Rearrange the formula so T2 is on the left
T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.
T2 = 162 * 3.6* 273 / (56 *1)
T2 = 159213.6/56
T2 = 2843.1
