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3 years ago

In the molecule NO2 which element is pulling the electrons

Chemistry

1 answer:

Andrei [34K]3 years ago

7 0

The bond between the N and 0 (double bond) transfers and gives a -ve charge on O and a +ve charge on N atom at the group . Thus the +vely charged nitrogen is electron-deficient pulling electrons towards itself!

The combination of the +vely charged nitrogen and the electronegative oxygen atom leads to delocalization causing the resonance effect.

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An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f

Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

Mass of carbon, $C$ :

$\frac{32.18}{100}\times 223.94 = 72.06 g$

Mass of hydrogen, $H$ :

$\frac{4.5}{100}\times 223.94 = 10.08 g$

Mass of chlorine, $Cl$ :

$\frac{63.32}{100}\times 223.94 = 141.79 g$

Now, the number of each element in the unknown compound is determined by the formula:

$number of moles = \frac{given mass}{molar mass}$

Number of moles of $C$ :

$number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole$

Number of moles of $H$ :

$number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole$

Number of moles of $Cl$

$number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole$

Dividing each mole with the smallest number of mole, to determine the empirical formula:

$C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}$

$C_{1.5}H_{2.5}Cl_{1}$

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is $C_{3}H_{5}Cl_{2}$ .

Empirical mass = $12\times 3+1\times 5+2\times 35.5 = 112 g/mol$

In order to determine the molecular formula:

n = $\frac{molar mass}{empirical mass}$

n = $\frac{223.94}{112} = 1.99 \simeq 2$

So, the molecular formula is:

$2\times C_{3}H_{5}Cl_{2} = C_{6}H_{10}Cl_{4}$

6 0

3 years ago

How many grams of CO2 gas are produced if 48 grams of CH4 gas are consumed?

NNADVOKAT [17]

Answer:

CH4 +2 O2 — CO2 +2 H2O

Now we see that for 1 mol i.e. 16 grams of methane results in 1 mol of CO2 or 44 grams of CO2.

That means for 3 moles of methane , we will obtain 3 moles of CO2 OR for 48 (3*16) grams of CO2 , we will obtain (44*3) 132 grams of CO2 .That's it….

Explanation:

hey .dude hope the answer was helpful ....

5 0

2 years ago

What part of the atom contains the most space: the nucleus or the energy levels? Why?

Anarel [89]

Answer:

I don't know 100% but im pretty sure its electrons, if im wrong im really sorry let me know in the comments ill change it

explanation

the nucleus has more weight but its more compact but the electrons are spread apart circling the nucleus and therefor take up more space

3 0

2 years ago

Oxalic acid can remove rust (Fe2O3) caused by bathtub rings according to the reaction Fe2O3(s) - 6H2C2O4(aq) rightarrow 2Fe(C2O4

Katena32 [7]

Answer: The mass of rust that can be removed is 1.597 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of oxalic acid solution = 0.1255 M

Volume of solution = $6.00\times 10^2mL$ = 600 mL = 0.600 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of oxalic acid}}{0.600L}\\\\\text{Moles of oxalic acid}=(0.100mol/L\times 0.600L)=0.06mol$

For the given chemical reaction:

$Fe_2O_3(s)+6H_2C_2O_4(aq.)\rightarrow 2Fe(C_2O_4)_3^{3-}(aq.)+3H_2O(l)+6H^+(aq.)$

By Stoichiometry of the reaction:

6 moles of oxalic acid reacts with 1 mole of ferric oxide (rust)

So, 0.06 moles of oxalic acid will react with = $\frac{1}{6}\times 0.06=0.01mol$ of ferric oxide (rust)

To calculate the mass of rust for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of rust (ferric oxide) = 159.7 g/mol

Moles of rust = 0.01 moles

Putting values in above equation, we get:

$0.01mol=\frac{\text{Mass of rust}}{159.7g/mol}\\\\\text{Mass of rust}=(0.01mol\times 159.7g/mol)=1.597g$

Hence, the mass of rust that can be removed is 1.597 grams

5 0

3 years ago

WILL MARK BRAINLIEST

levacccp [35]

Okay so i think to calculate volume change you use the ideal gas law . this is pressure x volume = amount of substance x ideal gas constant x temperature. i honestly don’t know where to go from there but i hope this helped a bit :(

8 0

3 years ago