Answer:
Eukarya- Multicellar and unicellar
Explanation:
Fugi- Multicellar
Protista- unicellar
eubacteria- unicellar and have no nucleus
Eukarya is the only option, and that is because Eukarya is both! It contains a nucleus, and a membrane making it the only one that fits into the incomplete graphic! Hope this helps! (I noticed you were waiting for a while so I came to help! <3)
The more acid the salt, the lower the pH. So we must find the salt that is most basic among the options. Our answer is
![NaCH_3COO](https://tex.z-dn.net/?f=NaCH_3COO)
, because it is formed by a weak acid (
![H_3COOH](https://tex.z-dn.net/?f=H_3COOH)
) with a strong base (
![NaOH](https://tex.z-dn.net/?f=NaOH)
).
The reason why Tin granules is mostly used or preferred than
just a piece of Tin because a piece of Tin is plainly metal, in which this will
be difficult in having it to be oxidized when it is exposed or in the air
rather than Tin granules.
Answer:
In Cl
, the 2 is a subscript because it indicates there are 2 of the same elements. The Lewis structure would display it as Cl-Cl.
On the other hand, a superscript would indicate a specific charge.
All subscripts show the amount of the specific element there is.
An example would be O
or N
, they both show that there are 2 of the same elements.
If the subscript is outside a parenthesis such as
it indicates there are 2
molecules.
Hi! the english version of the given question is "A helium balloon is inflated to the volume of 0.045 m3 at a temperature of 2 ° C. If the balloon is cooled to -12 ° C. What will its new volume be? Consider that the pressure does not vary". The answer is given in english.
Answer:
The new volume of balloon is
.
Explanation:
Let's assume the helium gas inside the balloon behaves ideally.
The total number of moles of helium gas inside the balloon remains constant.
Applying combined gas law, we get:
![\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7DV_%7B1%7D%7D%7BT_%7B1%7D%7D%3D%5Cfrac%7BP_%7B2%7DV_%7B2%7D%7D%7BT_%7B2%7D%7D)
Where:
are the initial and final pressure of the balloon respectively.
are the initial and final volume of the balloon respectively.
are the initial and final temperature in the kelvin scale respectively.
Given:
![P_{1}=P_{2}](https://tex.z-dn.net/?f=P_%7B1%7D%3DP_%7B2%7D)
![V_{1}=0.045\;m^{3}](https://tex.z-dn.net/?f=V_%7B1%7D%3D0.045%5C%3Bm%5E%7B3%7D)
![T_{1}=(273+2)\;K=275\;K](https://tex.z-dn.net/?f=T_%7B1%7D%3D%28273%2B2%29%5C%3BK%3D275%5C%3BK)
![T_{2}=(273-12)\;K=261\;K](https://tex.z-dn.net/?f=T_%7B2%7D%3D%28273-12%29%5C%3BK%3D261%5C%3BK)
Substituting the above values, we get:
![\frac{(0.045\;m^{3})}{275\;K}=\frac{V_{2}}{261\;K}](https://tex.z-dn.net/?f=%5Cfrac%7B%280.045%5C%3Bm%5E%7B3%7D%29%7D%7B275%5C%3BK%7D%3D%5Cfrac%7BV_%7B2%7D%7D%7B261%5C%3BK%7D)
![\Rightarrow V_{2}=0.043\;m^{3}](https://tex.z-dn.net/?f=%5CRightarrow%20V_%7B2%7D%3D0.043%5C%3Bm%5E%7B3%7D)