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RUDIKE [14]
3 years ago
11

Which accurately labels the Golgi body?

Chemistry
1 answer:
dexar [7]3 years ago
4 0

Answer:Z

Explanation:

Just took the quiz

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Equivalencia de un año luz en SI
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Hi uhh i've never really typed in Spanish but here I go! :)

 Un ano luz es equivalente a 9.467 PM
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3 years ago
Part A The first step to engineering is to define the problem. Write down the problem the students have to solve, and describe t
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Answer:

Engineering is all about solving problems using math, science, and technical knowledge. And engineers have solved a lot of problems in the world by designing and building various technologies. We have everything from machines that can breathe for you in hospitals to suspension bridges to computers we use every day. All of these things were once designed by engineers using the engineering design process.

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A piece of solid carbon dioxide with a mass of 5.50 grams is placed in a 10 Liter vessel that already contains 705 torr at 24 Ce
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0.58 mol of Mg contains how many atoms? please show work
Jet001 [13]

The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms

<h3>Avogadro's hypothesis </h3>

1 mole of Mg = 6.02×10²³ atoms

<h3>How to determine the atoms in 0.58 mole of Mg </h3>

1 mole of Mg = 6.02×10²³ atoms

Therefore,

0.58 mole of Mg = 0.58 × 6.02×10²³

0.58 mole of Mg = 3.49×10²³ atoms

Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg

Learn more about Avogadro's number:

brainly.com/question/26141731

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3 0
2 years ago
Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.
Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}

Converting moles of boric acid to mass:

0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}}   = 6.183 g

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.


5 0
3 years ago
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