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3 years ago

How many grams are in 6 moles of MgCl2?

Chemistry

1 answer:

maks197457 [2]3 years ago

3 0

First, you have to refer back to your table of elements.

Break it down. When they say how many grams, it means the atomic weight. In this case, we have a chemical compound.

MgCl2 mass:

Mg= 24.305

Cl2= 35.453 x 2= 70.906

Mg + Cl2= 24.305 + 70.906 = 95.211

So, one mole of MgCl2 is 95.211 grams.

To get what you are looking for, multiply this by 6...

95.211 x 6 =…

I made it pretty simple for you, all you have to do is multiply.

Hope I helped!

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What is the mass in grams of 1.00 x 10 24 atoms of Mn?

Alex17521 [72]

<h3>Answer:</h3>

91.2 g Mn

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

<u>Step 2: Identify Conversions</u>

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 91.2321 \ g \ Mn$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

91.2321 g Mn ≈ 91.2 g Mn

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3 years ago

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___________ a mixture of iron filings and sulfur will make the iron and sulfur atoms combine chemically

musickatia [10]

Iron sulfide is a<span> mixture of iron filings and sulfur will make the iron and sulfur atoms combine chemically.

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2 years ago

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200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486

ololo11 [35]

The empirical formula of a given compound is C6H9ON5.

<u>Explanation</u>:

Step 1: Obtain the mass of each element present in grams

Element % = mass in g = m

Carbon = 83.884 grams, Hydrogen = 10.486 grams, Oxygen = 18.640 grams, Nitrogen = 86.99 grams.

Step 2: Determine the number of moles of each type of atom present

m/atomic mass = Molar amount (M)

Molar amount of carbon = (83.884 1 mol ) / 12 g = 6.99

Molar amount of hydrogen = (10.486 1 mol) / 1 g = 10.49

Molar amount of oxygen = (18.64 1 mol) / 16 g = 1.17

Molar amount of nitrogen = (86.99 1 mol) / 14 g = 6.21

Step 3: Divide the number of moles of each element by the smallest number of moles

M / least M value = Atomic Ratio (R)

Atomic radius of carbon = 6.99 / 1.17 = 5.9 = 6

Atomic radius of hydrogen = 10.49 / 1.17 = 8.9 = 9

Atomic radius of oxygen = 1.17 / 1.17 = 1

Atomic radius of nitrogen = 6.21 / 1.17 = 5

Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

R * whole number = Empirical Formula

The empirical formula of a given compound is C6H9ON5.

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3 years ago

An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not

Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

$5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} )$ = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

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2 years ago

2) Which liberates the most energy?

evablogger [386]

Answer: $F(g)+e^-\rightarrow F^-(g)$

Explanation:

Electron gain enthalpy is defined as energy released on addition of electron to an isolated gaseous atom.

The amount of energy released will be maximum when the tendency to attract electrons is maximum. As flourine has atomic number of 9 and has electronic configuration of 2,7. It can readily gain 1 electron to attain stable noble gas configuration and hence liberates maximum energy.

$F(g)+e^-\rightarrow F^-(g)$

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3 years ago