Answer:
<h2>0.52 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
volume = final volume of water - initial volume of water
volume = 35 - 8 = 27 mL
We have
We have the final answer as
<h3>0.52 g/mL</h3>
Hope this helps you
The answer to your question is Hubble’s law
Answer: The correct option is heterogeneous mixture whose components are attracted differently to a magnet.
Explanation: There are two types of mixtures:
1) Homogeneous mixtures: In these mixtures, the particles are uniformly distributed throughout the mixture. These particles cannot be separated.
2) Heterogeneous Mixtures: These are the mixtures where the particles are visible separated and are not-uniformly distributed. These particles can be separated easily.
If magnet is used to separate the components of a mixture, the heterogeneous mixtures will only get separated.
To separate the components by a magnet, the components of a mixture should attract the magnet differently. One component should attract the magnet and another should not. Hence, they can be easily separated.
Hey there! Let's get that problem solved!
First: Let's define, "solution."
Solution: <span>a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).
Next: Ask yourself, "can a solution be taken apart?"
In some cases, yes. It can.
The solution of salt water for example, can be physically separated by evaporation. (place salt-water in a pot on a heated stove, place the cover to the pot on the opening, wait a few minutes, remove the top, and you can (and taste) the water without the salt!) </span><span />
Answer:
Explanation:
From the information given:
no of moles of 
= 0.01 L × 0.0010 mol/L
no of moles of = 
no of moles of 
= 0.01 L × 0.00010 mol/L
no of moles of = 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2
