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3 years ago

A particular element has the following set of properties:

Chemistry

1 answer:

natita [175]3 years ago

7 0

Answer:

D. bromine (Br)

Explanation:

Bromine is an element of halogen group.

Halogens, or "salt-formers," are located in group 17 of the periodic table. The halogens have the following properties:

1) They have low melting points and low boiling points.

2) They exist as diatomic molecules in the gas phase.

3) They have seven electrons in their outermost shell, which means they have seven valence

electrons.

4) They are highly reactive nonmetals.

5) They tend to gain electrons and form negative ions.

Thus the given properties are of Bromine.

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A chemistry student was asked to calculate the number of moles of iron required to react with 1.20 mol of oxygen to produce iron

nikitadnepr [17]

I’m on the same one smh

6 0

2 years ago

A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.

Bingel [31]

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

$n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH$

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

$m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO$

And consequently the moles:

$n_O=0.12molO$

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

$C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1$

Thus, the empirical formula, taken the nearest whole number is:

$C_5H_{10}O_2$

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

Best regards!

6 0

2 years ago

Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi

ivann1987 [24]

Answer:The formulas of ionic compounds are:

a) $CuBr$

b) $Mn_2O_3$

c) $Hg_2I_2$

d) $Mg_3(PO_4)_2$

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains $Cu^+$ ion.

$Cu^++Br^-\rightarrow CuBr$

(b) Manganese oxide : Given that it contains $Mn^{3+}$ ion.

$Mn^{3+}+O^{2-}\rightarrow Mn_2O_3$

(c)Mercury iodide :Given that it contains $Hg_2^{2+}$

$Hg_2^{2+}+I^-\rightarrow Hg_2I_2$

(d) Magnesium phosphate :Given that it contains $PO_4^{3-}$

$Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2$

4 0

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!

6 0

2 years ago

Love yall hope you doing good also anyone feel free to talk

statuscvo [17]

I’m doing good thanks

5 0

2 years ago

Read 2 more answers