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Strike441 [17]
3 years ago
7

When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat

ionship between those units so you will know how to set up your equation in order to solve the problem. Review the data sets below and use the steps of the problem-solving method to determine whether the given measurements would be appropriate for calculating mass, volume, or density.
a. 432 g of table salt occupies 20.0 cm^3 of space
b. 5.00 g 0T balsa wood, density of balsa wood : 0.16 g/cm^3
c. 32 cm^3 sample of gold density of gold 19.3 =g/cm^3
d. 150 g of iron, density of Iron = 79.0 g/cm^3
Chemistry
1 answer:
UNO [17]3 years ago
5 0

Answer:

See Explanation

Explanation:

Given

(a) to (d)

Required

Determine whether the given parameters can calculate the required parameter

To calculate either Density, Mass or Volume, we have

Density = \frac{Mass}{Volume}

Mass = Density * Volume

Volume = \frac{Mass}{Density}

(a) 432 g of table salt occupies 20.0 cm^3 of space

Here, we have:

Mass = 432g

Volume = 20.0cm^3

The above can be used to calculate Density as follows;

Density = \frac{Mass}{Volume}

Density = \frac{432g}{20.0cm^3}

Density = 21.6g/cm^3

(b) 5.00 g of balsa wood, density of balsa wood : 0.16 g/cm^3

Here, we have:

Mass = 5.00g

Density = 0.16g/cm^3

This can be used to solve for Volume as follows:

Volume = \frac{Mass}{Density}

Volume = \frac{5.00g}{0.16g/cm^3}

Volume = 31.25cm^3

(c) 32 cm^3 sample of gold density of 19.3 g/cm^3

Here, we have:

Volume = 32cm^3

Density = 19.3g/cm^3

This can be used to calculate Mass as follows:

Mass = Density * Volume

Mass = 32cm^3 * 19.3g/cm^3

Mass = 617.6g

(d) 150 g of iron, density of Iron = 79.0 g/cm^3

Here, we have

Mass = 150g

Density = 79.0g/cm^3

This can be used to calculate volume as follows:

Volume = \frac{Mass}{Density}

Volume = \frac{150g}{79.0g/cm^3}

Volume = 1.90 cm^3 <em>Approximated</em>

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Hence, the correct answer is "Storage battery or cell".

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1 year ago
Help asap. <br>How many moles are 5.55x10^25 atoms of Mg
LekaFEV [45]

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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