Answer:
They both go backward because of force.
Explanation:
The logic behind this answer is that child right and pushes away causing her to go backward meaning her partner is being pushed backwards to.
(Hope this was helpful!)
<span>AX(aq)+BY(aq)→no precipitate
AX(aq)+BZ(aq)→precipitate
this two equations imply
</span>
AX(aq) is soluble and <span>BY(aq) is insoluble
the answer is
</span><span>E. BY</span>
Answer:
No
Explanation:
The mass fraction is defined as:

where:
- wi: mass fraction of the substance i
- mi: mass of the substance i
- mt: total mass of the system
<u><em>The mass fraction of two substances (A and B), will be the same, ONLY if the mass of the substance A (mA) is the same as the mass of the substance B (mB).</em></u>
An equimolar mixutre of O2 and N2 has the same amount of moles of oxygen and nitrogen, just to give an example let's say that the system has 1 mole of O2 and 1 mole of N2. Then using the molecuar weigth of each of them we can calculate the mass:
mA= 1 mole of O2 * 16 g/1mol = 16 g
mB=1 mole of N2 *28 g/1mol=28 g
As mA≠mB then the mass fractions are not equal, so the answear is NO.
The dissociation of formic acid is:

The acid dissociation constant of formic acid,
is:
![k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}](https://tex.z-dn.net/?f=%20k_a%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%20%20%5BH%5E%7B%2B%7D%5D%7D%7BHCOOH%7D%20%20%20%20%20)
Rearranging the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
pH = 2.75
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D%20)
![[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ctimes%2010%5E%7B-3%7D%20)


Substituting the values in the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7B1.78%5Ctimes%2010%5E%7B-4%7D%7D%7B1.78%5Ctimes%2010%5E%7B-3%7D%7D%20%20%20)
Hence, the ratio is
.