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3 years ago

What happened to the wind speeds of the storms?

Chemistry

1 answer:

oksano4ka [1.4K]3 years ago

6 0

They would most likely speed up.

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What would the formula be for this model above?

Nadusha1986 [10]

Probably CH(subscript)4... :) It's Methane

4 0

3 years ago

What is the bond energy required to break one mole of carbon-carbon bonds

ArbitrLikvidat [17]

Answer:

<h2><u><em>100 kcal of bond energy</em></u></h2>

8 0

3 years ago

How many grams of KNO3 are required to prepare 0.250 L of 0.70 M solution?

drek231 [11]

Answer:

about 3

Explanation:

just did this for the points just being honiest

4 0

2 years ago

Explain the electrolysis of acidified water

Andrew [12]

Answer:

yqaeh

Explanation:

Electrolysis of acidified water

Water is a poor conductor of electricity, but it does contain some hydrogen ions , H +, and hydroxide ions, OH -. These ions are formed when a small proportion of water molecules naturally dissociate . ... H + ions are attracted to the cathode , gain electrons and form hydrogen gas.

8 0

2 years ago

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

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:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$

Now, we compute the total moles of bromide:

$n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol$

Then, the total volume in liters:

$150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L$

Therefore, the concentration of total bromide is:

$M=\frac{0.12338mol}{0.325L}\\\\M=0.380M$

Best regards!

8 0

2 years ago