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GenaCL600 [577]

3 years ago

12

a student balanced the chemical equation for the reaction of magnesium with oxygen by writing Mg + O2 ----> MgO2. What is wro

ng with this equation?

1 answer:

egoroff_w [7]3 years ago

6 0

Answer:

Explanation:

Magnesium when it oxidizes has a valence of 2.

Oxygen, when it mixes with something, has a valence of - 2

So Mg and O2 will form something, but what? The answer is MgO

Mg + O2 ===> MgO

To balance this equation, you need 2 Magnesiums on the right and 2 Oxygens also on the right. The left will need a two Magnesiums.

The balanced equation will be

2Mg + O2 ====> 2MgO

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2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

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3 years ago

Emily learns that ancient astronomers, such as Ptolemy, created a model of the solar system based only on what they could see wi

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Explanation:

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2. Answer the following questions about a sample of calcium phosphate:

weqwewe [10]

Answer:

a) <u>310.18 g/mol</u>

<u>b) 4.352 moles Ca3(PO4)2</u>

<u>c) 2.6 * 10^24 molecules</u>

<u>d) 5.24 * 10^24 P atoms</u>

<u>e)13.056 moles Ca</u>

<u>f)</u>10825.3 grams

Explanation:

Step 1: Data given

Atomic mass of Ca = 40.08 g/mol

Atomic mass of P = 30.97 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molecular weight of Ca3(PO4)2

Molecular weight of Ca3(PO4)2 = 3*atomic mass of Ca + 2* atomic mass of P and 8* atomic mass of O

Molecular weight of Ca3(PO4)2 = 3*40.08 + 2*30.97 + 8*16.0 =<u> 310.18 g/mol</u>

Step 3: Calculate moles of Ca3(PO4)2 in 1350 grams

Moles Ca3(PO4)2 = mass Ca3(PO4)2 /molar mass

Moles Ca3(PO4)2 = 1350 grams / 310.18 g/mol

Moles Ca3(PO4)2 = <u>4.352 moles</u>

Step 4: Calculate molecules in 1350 grams

Molecules = moles * number of Avogadro

Molecules = 4.352 moles * 6.02 * 10^23

Molecules = <u>2.6 *10^24 molecules</u>

<u />

Step 5: Calculate moles Phosphorus

For 1 mol Ca3(PO4)2 we need 2 moles P

For 4.352 moles Ca3(PO4)2 we have 2*4.352 = 8.704 moles

Step 6: Calculate P atoms

Atoms P = 8.704 moles * 6.02*10^23

Atoms P =<u> 5.24 * 10^24 P atoms</u>

<u />

Step 7: Calculate moles Calcium in 1350 grams

For 1 mol Ca3(PO4)2 we have 3 moles Ca

For 4.352 moles we have 3*4.352 = <u>13.056 moles Ca</u>

<u />

<u />

<u>Step 8:</u> Calculate mass of 2.1 * 10^25 molecules of Ca3(PO4)2

Moles Ca3(PO4)2 = 2.1 * 10^25 / 6.02 * 10^23

Moles Ca3(PO4)2 = 34.9 moles

Mass Ca3(PO4)2 = 34.9 moles * 310.18 g/mol

Mass Ca3(PO4)2 = 10825.3 grams

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3 years ago