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sineoko [7]
3 years ago
5

How is the name of the second element in a covalent molecule changed

Chemistry
2 answers:
Aleonysh [2.5K]3 years ago
8 0

Answer:

See explanation

Explanation:

Here we are trying to see how a binary covalent compound is named. A binary covalent compound comprises of only two elements held together by covalent bonds.

The first element retains its normal name whereas the second element has the suffix -ide added to it.

For instance, CO2 is named as carbon dioxide, HBr is named as hydrogen bromide etc.

Alex Ar [27]3 years ago
8 0

Answer: it's ide

Explanation:

Got it right in the quiz

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a

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Sarin is a deadly nerve gas that was outlawed by the Chemical Weapons Convention of 1993. However, it has been used at various t
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Answer:

The correct answer is - a.  Fluorine is the leaving group in Sarin.

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3 years ago
Pure nitrogen (N2) and pure hydrogen (H2) are fed to a mixer. The product stream has 40.0% mole nitrogen and the balance hydroge
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Explanation:

The given data is as follows.

        Mass flow rate of mixture = 1368 kg/hr

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This means that H_{2} in feed = (100 - 40)% = 60%

We assume that there are 100 total moles/hr of gas (N_{2} + H_{2}) in feed stream.

Hence, calculate the total mass flow rate as follows.

           40 moles/hr of N_{2}/hr (28 g/mol of N_{2}) + 60 moles/hr of H_{2}/hr (2 g/mol of H_{2})

                  40 \times 28 g/hr + 60 \times 2 g/hr    

                  = 1120 g/hr + 120 g/hr

                  = 1240 g/hr

                  = \frac{1240}{1000}              (as 1 kg = 1000 g)

                  = 1.240 kg/hr

Now, we will calculate mol/hr in the actual feed stream as follows.

                 \frac{100 mol/hr}{1.240 kg/hr} \times 1368 kg/hr

                   = 110322.58 moles/hr

It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of N_{2} into the reactor as follows.

                       0.4 \times 110322.58 moles/hr

                      = 44129.03 mol/hr

As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.

Therefore, calculate the rate flow of N_{2} into the reactor as follows.

                       0.028 kg \times 44129.03 mol/hr

                         = 1235.612 kg/hr

Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.

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Adding the two equations would yield to:

FePO4(s) + H3O+(aq) ⇌ Fe^3+(aq) + HPO4^2−(aq) + H2O(l)
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