Answer:
6782 has 4, 26.2 has 3, 0.3491 has 4 and 55 has 2 significant figures .
Answer:
M=0.380 M.
Explanation:
Hello there!
In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:
![n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-](https://tex.z-dn.net/?f=n_%7BBr%5E-%7D%5E%7Bin%5C%20AlBr_3%7D%3D4.85%20gAlBr_3%2A%5Cfrac%7B1molAlBr_3%7D%7B266.69gAlBr_3%7D%2A%5Cfrac%7B3molBr%5E-%7D%7B1molAlBr_3%7D%20%20%3D0.05456molBr%5E-%5C%5C%5C%5Cn_%7BBr%5E-%7D%5E%7Bin%5C%20ZnBr_2%7D%3D7.75gZnBr_2%2A%5Cfrac%7B1molZnBr_2%7D%7B225.22gZnBr_2%7D%2A%5Cfrac%7B2molBr%5E-%7D%7B1molZnBr_2%7D%20%20%3D0.06882molBr%5E-)
Now, we compute the total moles of bromide:
![n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol](https://tex.z-dn.net/?f=n_%7BBr%5E-%7D%3D0.05456mol%2B0.06882mol%5C%5C%5C%5Cn_%7BBr%5E-%7D%3D0.12338mol)
Then, the total volume in liters:
![150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L](https://tex.z-dn.net/?f=150mL%2B175mL%3D325mL%2A%5Cfrac%7B1L%7D%7B1000mL%7D%20%5C%5C%5C%5C%3D0.325L)
Therefore, the concentration of total bromide is:
![M=\frac{0.12338mol}{0.325L}\\\\M=0.380M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B0.12338mol%7D%7B0.325L%7D%5C%5C%5C%5CM%3D0.380M)
Best regards!
I think the answers to this is d
Answer:
Light waves are electromagnetic waves while sound waves are mechanical waves. light waves are transverse while sound waves are longitudinal. sound waves require a material medium to travel, and hence, cannot travel in a vacuum. the speed of light in a medium is constant.
Explanation:
Hope this helped Mark BRAINLIEST!!
I believe that number 27 is A