Answer:
2.01 x 103 mm Hg
Explanation:
To convert from atmospheres to millimeters of mercury, it is necessary to know the equivalence between the two units (1 atm = 760.0 mm Hg). The atmosphere units should be arranged to cancel out in order to obtain units in millimeters of mercury.
Answer:
The centripetal acceleration of the object is 
.
Explanation:
We have,
Radius of a circular path is 2 m
It takes 5 seconds to complete 10 revolutions. It means its angular velocity is :
The centripetal acceleration of the object is given by :
So, the centripetal acceleration of the object is .
The solution is the mixture of the solute into the solvent particles. In 60 mL solution, 59.8 mL of solvent is added.
<h3>What are solute and solvent?</h3>
Solute and solvent are the components of a solution. They react and mix together to produce a homogenous mixture. The solute is a substance that is added to the solvent.
Solvents are substances that dissolve the added solute in them to make a homogenous solution.
Given,
Volume of Solute = 0.20 mL
Volume of Solution = 60 mL
The volume of solvent is calculated as:
Volume of solvent = Volume of solution - Volume of solute
= 60- 0.20
= 59.8 mL.
Therefore, 59.8 mL of solvent is present in 60 mL of solution.
Learn more about solute and solvent here:
brainly.com/question/20458032
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Explanation:
According to mass action,
![\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2](https://tex.z-dn.net/?f=%5Ctextrm%7Brate%7D%3D-%5Cdfrac%7B%5CDelta%5B%5Ctextrm%20A%5D%7D%7B2%5CDelta%20t%7D%3Dk%5B%5Ctextrm%20A%5D%5E2)
Where, k is the rate constant
So,
![\dfrac{d[A]}{dt}=-k[A]^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-k%5BA%5D%5E2)
Integrating and applying limits,
![\int_{[A_t]}^{[A_0]}\frac{d[A]}{[A]^2}=-\int_{0}^{t}kdt](https://tex.z-dn.net/?f=%5Cint_%7B%5BA_t%5D%7D%5E%7B%5BA_0%5D%7D%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%5E2%7D%3D-%5Cint_%7B0%7D%5E%7Bt%7Dkdt)
we get:
![\dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20%5Cdfrac%7B1%7D%7B%5BA%5D_0%7D%2Bkt)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Half life is the time when the concentration reduced to half.
So, ![[A_t]=\frac{1}{2}\times [A_0]](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5BA_0%5D)
Applying in the equation as:
![t_{1/2}=\dfrac{1}{k[A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA_o%5D%7D)
To identify in what beakers will a reaction occur, we need to know what aqueous solution is in the four beakers. Then, we also need to know what metals are placed inside those beakers. Now, we can analyze which metals might or will react to the solution in the beakers.
