Solution. because the substances can't go back to their original form. like kool aid, when its mixed you cant separate the powder and the water again.
Explanation:
Climate is the period of time average weather condition which occurs at a place
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
Hello,
Here is your answer:
The proper answer to this question is that "e<span>ach sub-level electron type has a unique path where it will likely to be found".
If you need anymore help feel free to ask me!
Hope this helps!</span>
Answer:
![m_{H_2O}=3.384gH_2O](https://tex.z-dn.net/?f=m_%7BH_2O%7D%3D3.384gH_2O)
Explanation:
Hello,
In this case, the chemical reaction is:
![Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O](https://tex.z-dn.net/?f=Cd%28OH%29_2%2B2HBr%5Crightarrow%20CdBr_2%2B2H_2O)
Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:
![n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O](https://tex.z-dn.net/?f=n_%7BH_2O%7D%5E%7Bby%5C%20Cd%28OH%29_2%7D%3D19.3gCd%28OH%29_2%2A%5Cfrac%7B1molCd%28OH%29_2%7D%7B146.4gCd%28OH%29_2%7D%2A%5Cfrac%7B2molH_2O%7D%7B1molCd%28OH%29_2%7D%3D0.264molH_2O%5C%5C%5C%5Cn_%7BH_2O%7D%5E%7Bby%5C%20HBr%7D%3D15.21gHBr%2A%5Cfrac%7B1molHBr%7D%7B80.9gHBr%7D%2A%5Cfrac%7B2molH_2O%7D%7B2molHBr%7D%3D0.188molH_2O)
In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:
![m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O](https://tex.z-dn.net/?f=m_%7BH_2O%7D%3D0.188molH_2O%2A%5Cfrac%7B18gH_2O%7D%7B1molH_2O%7D%5C%5C%20%5C%5Cm_%7BH_2O%7D%3D3.384gH_2O)
Regards.