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2 years ago

How many halogens are in the period 3 of the periodic table

2 answers:

8 0

Halogen, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts).

I hope this helps you.

Natasha_Volkova [10]2 years ago

8 0

Answer: 17

Explanation:

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What element is likely to form an ionic compound with chlorine?

kirill [66]

Sodium (NA)
the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.

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2 years ago

The mass spectrum of an unknown compound has a molecular ion peak with a relative intensity of 57.10% and an M+1 peak of 6.83%.

max2010maxim [7]

There are 11 Carbon atoms in the compound.

<u>Solution:</u>

Carbon atom count is the ratio of the M peak to the M+1 peak.

$\text{ Number of Carbon atoms }=\frac{\text { Relative intensity of } M+1 \text { peak }}{0.011 \times \text { Relative tntensity of } M \text { peak }}$

Here M peak is 57.10% and M+1 peak is 6.83%. On applying the values in the formula we get,

$\frac{0.0683}{0.011\times0.571g} = 10.87\approx11$

Therefore, the number of Carbon atoms in the compound are 11.

Refer the image attached below for a better understanding of M peak and M+1 peak.

The heaviest ion that has the greatest m/z value is said to be the molecular ion peak in mass spectrum.

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2 years ago

How many grams of octane (C8H18) must be burned to produce 300.0g of CO2?

nika2105 [10]

Answer:

$m_{C_8H_{18}}=85.67gC_8H_{18}$

Explanation:

Hello there!

In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:

$m_{C_8H_{18}}=300.0gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molC_8H_{18}}{25molO_2}*\frac{114.22gC_8H_{18}}{1molC_8H_{18}} \\\\m_{C_8H_{18}}=85.67gC_8H_{18}$

Regards!

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