Answer and Explanation: Sample A is a <u>mixture</u>. Solubility is characteristics of each substance, which means a substance can be distinguished from other substances and can be useful to separate mixtures.
In Sample A, when is added different volumes of water, the resulting powder has different mass. This means there are more than one substance forming the yellow cube. Therefore, sample A is a mixture.
Sample B is a <u>pure</u> <u>substance</u>. Each substance has its own melting point. Whe na pure substance reaches its melting point, temperature is constant until all of that substance is melted. In sample B, temperature is stable at 66.2°C and then, after all the powder is melted, it rises again. Therefore, sample B is a pure substance.
Answer: Mothballs have weak intermolecular forces.
No all substances do not behave like mothballs at normal conditions. Example: benzene , chloroform
Explanation:
Sublimation is a process of converting a substance from solid state to gaseous state without the formation of liquid at constant temperature.
A substance which undergoes sublimation is called as sublimating substance.
As mothballs is made of napthalene which has weak inter molecular forces of attraction between its molecules, it directly sublimes into gaseous state without leaving any residue and is called as a sublimating substance.
Not all substances behave like mothballs at normal conditions. Example: benzene , chloroform
Answer:
2
Explanation:
1 is definitely wrong
density must be lower to atop another liquid
in 3 ,I consider the example of water and oil whi has a lower boiling point
We are given that the balanced chemical reaction is:
cacl2⋅2h2o(aq) +
k2c2o4⋅h2o(aq) --->
cac2o4⋅h2o(s) +
2kcl(aq) + 2h2o(l)
We known that
the product was oven dried, therefore the mass of 0.333 g pertains only to that
of the substance cac2o4⋅h2o(s). So what we will do first is to convert this
into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is
molar mass of cac2o4 plus the
molar mass of h2o.
molar mass cac2o4⋅h2o(s) = 128.10
+ 18 = 146.10 g /mole
moles cac2o4⋅h2o(s) =
0.333 / 146.10 = 2.28 x 10^-3 moles
Looking at
the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is
1:1, therefore:
moles k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles
Converting
this to mass:
mass k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles (184.24 g /mol) = 0.419931006 g
Therefore:
The mass of k2c2o4⋅<span>h2o(aq) in
the salt mixture is about 0.420 g</span>
The answer is 1/16.
Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1.
![(1/2)^{n} = x](https://tex.z-dn.net/?f=%20%281%2F2%29%5E%7Bn%7D%20%3D%20x)
,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample
2.
![t_{1/2} = \frac{t}{n}](https://tex.z-dn.net/?f=%20t_%7B1%2F2%7D%20%3D%20%5Cfrac%7Bt%7D%7Bn%7D%20)
where:
<span>
![t_{1/2}](https://tex.z-dn.net/?f=%20t_%7B1%2F2%7D%20)
- half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>
![t_{1/2}](https://tex.z-dn.net/?f=%20t_%7B1%2F2%7D%20)
= 2.5 min
We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
![t_{1/2} = \frac{t}{n}](https://tex.z-dn.net/?f=%20t_%7B1%2F2%7D%20%3D%20%5Cfrac%7Bt%7D%7Bn%7D%20)
,
</span>Then:
![n = \frac{t}{ t_{1/2} }](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7Bt%7D%7B%20t_%7B1%2F2%7D%20%7D%20)
⇒
![n = \frac{10 min}{2.5 min}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B10%20min%7D%7B2.5%20min%7D%20)
⇒
![n=4](https://tex.z-dn.net/?f=n%3D4)
<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>
![(1/2)^{n} = x](https://tex.z-dn.net/?f=%20%281%2F2%29%5E%7Bn%7D%20%3D%20x)
</span>⇒
![x=(1/2)^4](https://tex.z-dn.net/?f=x%3D%281%2F2%29%5E4)
<span>⇒
![x= \frac{1}{16}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B1%7D%7B16%7D%20)
</span>