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ryzh [129]
3 years ago
14

Help asap. How many moles are 5.55x10^25 atoms of Mg

Chemistry
1 answer:
LekaFEV [45]3 years ago
3 0

Answer:

\boxed {\boxed {\sf About \ 92.2 \ moles \ of \ Magnesium}}

Explanation:

To convert from atoms to moles, we must Avogadro's number:

  • 6.022*10²³

This number tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, it is atoms of magnesium in 1 mole.

  • 6.022 *10²³ atoms Mg / 1 mol mg

Now, let's set up an expression using:

\frac {6.022 * 10^ {23} \ atoms \ Mg}{ 1 \ mol \ Mg}

Multiply this by the given number of atoms.

5.55*10^{25} \ atoms \ Mg *\frac {6.022 * 10^ {23} \ atoms \ Mg}{ 1 \ mol \ Mg}

Flip the fraction so the atoms of magnesium can cancel.

5.55*10^{25} \ atoms \ Mg *\frac { 1 \ mol \ Mg }{6.022 * 10^ {23} \ atoms \ Mg}

5.55*10^{25} *\frac { 1 \ mol \ Mg }{6.022 * 10^ {23}}

Multiply and condense this into 1 fraction.

\frac { 5.55*10^{25} \ mol \ Mg }{6.022 * 10^ {23}}

92.1620724 \ mol \ Mg

The original measurement had 3 significant figures, so our answer must have that many as well. For the number we calculated, that is the tenth place. The 6 in the hundredth place tells us to round the 1 to a 2.

92.2 \ mol \ Mg

It is about <u>92.2 moles of magnesium.</u>

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Answer:

0.233682161601125

Explanation:

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2 years ago
A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of
Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B =  21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A:   M  = m1 *0.8662    O = m1 *0.1338

compound B:   M  = m2 *0.90666    O = m2 *0.09334

compound C:   M  = m3 *0.92832    O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO  / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO  / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO  / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338   m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48  ⇒ m2 = 227.74 g

0.92832m3 = 206.48  ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams  ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

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