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4 years ago

How to find the charge of an element

Chemistry

1 answer:

katrin2010 [14]4 years ago

7 0

Answer:

the atomic number is how you find the charge

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An atom has three full orbitals in its second energy level. How many electrons are present in the second energy level of the ato

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What elem eat is this? That has three orbitals? Usually the first orbital will contain only two electrons. The second orbital would contain four, the third would contain 16

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3 years ago

Refer to the image please!

Ludmilka [50]

Answer:

9. Electron Affinity

10. The second option....

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2 years ago

Complete the chart. Atomic Number: 10. 1s

Goshia [24]

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7 0

3 years ago

The electrolysis of water forms H2 and O2. 2H2O 2H2 + O2 What is the percent yield of O2 if 10.2 g of O2 is produced from the de

BigorU [14]

<u>Answer:</u> The percent yield of oxygen gas is 67.53 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of water = 17.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{17.0g}{18g/mol}=0.944mol$

For the given chemical equation:

$2H_2O\rightarrow 2H_2+O_2$

By Stoichiometry of the reaction:

2 moles of water produces 1 mole of oxygen gas

So, 0.944 moles of water will produce = $\frac{1}{2}\times 0.944=0.472moles$ of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.472 moles

Putting values in equation 1, we get:

$0.472mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.472mol\times 32g/mol)=15.104g$

To calculate the percentage yield of oxygen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of oxygen gas = 10.2 g

Theoretical yield of oxygen gas = 15.104 g

Putting values in above equation, we get:

$\%\text{ yield of oxygen gas}=\frac{10.2g}{15.104g}\times 100\\\\\% \text{yield of oxygen gas}=67.53\%$

Hence, the percent yield of oxygen gas is 67.53 %.

7 0

4 years ago

Read 2 more answers

General Lab Questions: (8 points) 5. A) What is the main problem with precipitating strontium ions from the town's water using p

Umnica [9.8K]

Answer:

1) high pH is required

2) other ions are precipitated along with the strontium ions

Explanation:

According to the solubility rules all phosphates are insoluble except those of sodium, potassium, and ammonium. This implies that strontium phosphate is insoluble in water. This explains why strontium ions can be precipitated from drinking water supply using phosphate. The main problem with the precipitation of strontium using phosphate is that it usually requires a high pH as the precipitation occurs under very alkaline conditions.

The main reason why the results may not be accurate is that other ions are precipitated along with the strontium such as calcium ions and magnesium ions. This may lead to inaccurate determination of the amount of strontium ions present.

7 0

3 years ago

Read 2 more answers