How do I convert 2.4 x 10^21 formula units of magnesium hydroxide to grams?

1 answer:

5 0

Are you taking chemistry at the same time? If not, this is a pretty hard question.

Facts

1. As a formula unit, Magnesium Hydroxide has a chemical formula of Mg(OH)2
2. That means for every Magnesium, there are 2 oxygens and 2 hydrogens
3. The weights of these components are found on a periodic table. Since every periodic table is different, I'm going to approximate the results.

Mg = 24
O = 16
H = 1

4. One mole of anything has 6.02 * 10^23 (in this case) formula units associated with it.

Step One
Find the number of grams in 1 formula unit using the numbers from the periodic table.

1 Mg = 1 * 24 = 24
2 O = 2 * 16 = 32
2 H = 2 * 1 = 2
Total = 58 grams

So
1 mole of Mg(OH)2 = 58 grams

1 mole / 6.02 * 10 ^23 formula units = x mole / 2.4 * 10 ^ 21 formula units
Cross multiply

1 mole * 2.4 * 10^21 formula units = x mole * 6.02 * 10^23 formula units.
Divide both sides by 6.02 * 10^23

1 * 2.4 * 10^21 / 6.02 * 10 ^ 23 = x
x = 0.003987 moles

Now do this again for grams
1 mole / 58 grams = 0.003987 mole / x grams Cross multiply
1 mole * x = 0.003987 * 58 grams
x = 0.2232 grams <<<<< Answer

You might be interested in

A) Determine which amounts of postage can be formed using just 3-cent and 10-cent stamps.

RUDIKE [14]

wait i will take screen shot please bran-list

8 0

3 years ago

A ball is thrown from a height of 38 meters with an initial downward velocity of 3 m/s

ANTONII [103]

<h2>Hello!</h2>

The answer is:

The ball will hit the ground after $t=2.47s$

Since we are given a quadratic function, we can calculate the roots (zeroes) using the quadratic formula. We must take into consideration that we are talking about time, it means that we should only consider positive values.

So,

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

We are given the function:

$h=-5t^{2}-3t+38$

Where,

$a=-5\\b=-3\\c=38$

Then, substituting it into the quadratic equation, we have:

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}=\frac{-(-3)+-\sqrt{(-3)^{2} -4*-5*38} }{2*-5}\\\\\frac{-(-3)+-\sqrt{(-3)^{2} -4*-5*38} }{2*-5}=\frac{3+-\sqrt{9+760} }{-10}\\\\\frac{3+-\sqrt{9+760} }{-10}=\frac{3+-\sqrt{769} }{-10}=\frac{3+-27.731}{-10}\\\\t1=\frac{3+27.731}{-10}=-3.073s\\\\t2=\frac{3-27.731}{-10}=2.473s$