Answer:
that looks pretty and also well NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in the both certain plenty of dust and gas. Both also have active star-forming sites.
Answer:
1.3 mL
Explanation:
First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:
1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.
It should have 2 significant figures, because 1.2kg has 2 and we are dividing.
Answer : The pH of 0.289 M solution of lithium acetate at 
is 9.1
Explanation :
First we have to calculate the value of 
.
As we know that,
where,
= dissociation constant of an acid = 
= dissociation constant of a base = ?
= dissociation constant of water = 
Now put all the given values in the above expression, we get the dissociation constant of a base.
Now we have to calculate the concentration of hydroxide ion.
Formula used :
![[OH^-]=(K_b\times C)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%28K_b%5Ctimes%20C%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
where,
C is the concentration of solution.
Now put all the given values in this formula, we get:
![[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%285.5%5Ctimes%2010%5E%7B-10%7D%5Ctimes%200.289%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
![[OH^-]=1.3\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.3%5Ctimes%2010%5E%7B-5%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
Now we have to calculate the pH.
Therefore, the pH of 0.289 M solution of lithium acetate at is 9.1
Answer:
Close to the calculated endpoint of a titration - <u>Partially open</u>
At the beginning of a titration - <u>Completely open</u>
Filling the buret with titrant - <u>Completely closed</u>
Conditioning the buret with the titrant - <u>Completely closed</u>
Explanation:
'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.
As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.
Answer:
Explanation:
Hello,
In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:
Thus, we solve for the final pressure P2 to obtain it as shown below:
Hence, we notice that the temperature doubles as well as the pressure.
Best regards.
