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2 years ago

Which element cannot be the central atom of a lewis structure?.

Chemistry

1 answer:

zalisa [80]2 years ago

5 0

Answer:

Hydrogen

Explanation:

Hydrogen can never be central atom despite its low electronegativity

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Copper is commonly mined as an ore with a variable percent composition of copper (II) sulfide. This ore is also sometimes referr

pantera1 [17]

Answer:

$m_{CuO}=93.6gCuO$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CuS+\frac{3}{2} O_2\rightarrow CuO+SO_2$

Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:

$m_{CuO}^{theoretical}=1.00kgCuS*\frac{1000gCuS}{1kgCuS} *\frac{12.5gCuS}{100gCuS} *\frac{1molCuS}{95.6gCuS} *\frac{1molCuO}{1molCuS} *\frac{79.5gCuO}{1molCuO} \\\\m_{CuO}^{theoretical}=103.95gCuO$

Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:

$m_{CuO}=103.95gCuO*0.9\\\\m_{CuO}=93.6gCuO$

Regards.

6 0

3 years ago

Compare protons with electrons.

Step2247 [10]

Proton:
Positively charged
Inside nucleus
Mass - 1

Electrons:
Negatively charged
Outside the nucleus
Mass - 1/2000

4 0

3 years ago

Read 2 more answers

Please help, thanks!

Anastasy [175]

You need to use the formula--> P1V1= P2V2 (Boyles's law)

P1= 14 bar
V1= 312 mL
P2= ?
V2= 652 mL

now we plug the values into the formula.

(14 x 312) = (P2x 652)

P2= (14 x 312)/ 652= 6.70 bar

5 0

3 years ago

What is the pOH of a 5.6 x 10-5 M solution of cesium hydroxide (CsOH)?

Rudiy27

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

7 0

3 years ago

In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

Neporo4naja [7]

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For Li

Given mass = 2.50 g

Molar mass of Li = 6.94 g/mol

Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles

Given: For $N_2$

Given mass = 2.50 g

Molar mass of $N_2$ = 28.02 g/mol

Moles of $N_2$ = 2.50 g / 28.02 g/mol = 0.08924 moles

According to the given reaction:

$6Li+N_2\rightarrow 2Li_3N$

6 moles of Li react with 1 mole of $N_2$

1 mole of Li react with 1/6 mole of $N_2$

0.3602 mole of Li react with $\frac {1}{6}\times 0.3602$ mole of $N_2$

Moles of $N_2$ that will react = 0.06 moles

Available moles of $N_2$ = 0.08924 moles

$N_2$ is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of $Li_3N$

1 mole of Li gives 2/6 mole of $Li_3N$

0.3602 mole of Li react with $\frac {2}{6}\times 0.3602$ mole of $Li_3N$

Moles of $Li_3N$ = 0.12

Molar mass of $Li_3N$ = 34.83 g/mol

Mass of $Li_3N$ = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

Theoretical yield = 4.18 g

5 0

3 years ago