Answer:
A velocity-time graph shows how velocity changes over time. The sprinter's velocity increases for the first 4 seconds of the race, it remains constant for the next 3 seconds, and it decreases during the last 3 seconds after she crosses the finish line.
Yes, Fe(s)+CuSO4→FeSO4(aq)+Cu(s) is a redox reaction. The reducing agent is Fe(s), while the oxidizing agent is CuSO4.
<h3>Oxidation-reduction reaction (redox)</h3>
Oxidation-reduction reaction is the type of reaction in which electrons are transferred from one element to another. A compound or an element either gains are looses an electron in redox reactions.
Fe(s)+CuSO4→FeSO4(aq)+Cu(s)
From the equation above, Fe(s) is the reducing agent because it gives away 2 electrons and undergoes oxidation reaction.
From the equation above, CuSO4 is an oxidizing agent because it gives oxygen to Fe(s) and undergoes a reduction reaction.
Therefore, the reducing agent is Fe(s), while the oxidizing agent is CuSO4.
Learn more about redox reactions here:
brainly.com/question/26263317
Answer:
3.23 atm
Explanation:
Use the equation P₁V₁=P₂V₂. P₁ and V₁ are initial pressure and volume. P₂ and V₂ are final pressure and volume. Solve for P₂.
(1.88 L)(3.81 atm) = (2.22 L)(P₂)
7.1628 L*atm = (2.22 L)(P₂)
P₂ = 3.226 atm
Answer;
Mass in gram/molecular mass=no.of molecule of Cl2 gas/Avogadro’s number(NA) - - - - - - - - -(1)
Here,
Molecular mass of cl2=35.5*2=71
No.of cl2 molecule=5*10^22
Avogadro’s number(NA)=6.032*10^23
Now,substituting these all value in equation one,we get,
Mass in gram =(no.of molecule)/NA)*molecular mass
Or,Mass in gram =(5*10^22/6.023*10^23)*71
Or,Mass in gram = 5.89gm.Ans
Explanation:
The reaction equation will be as follows.
Calculate the amount of 
dissolved as follows.
It is given that 
= 0.032 M/atm and 
= 
atm.
Hence, ![[CO_{2}]](https://tex.z-dn.net/?f=%5BCO_%7B2%7D%5D)
will be calculated as follows.
= 
= 
= 
or, = 
It is given that 
As, ![K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
![4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}](https://tex.z-dn.net/?f=4.46%20%5Ctimes%2010%5E%7B-7%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B0.608%20%5Ctimes%2010%5E%7B-5%7D%7D)
![[H^{+}]^{2}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%5E%7B2%7D)
= 
![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
= 
Since, we know that pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
So, pH = 
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.
