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3 years ago

How many atoms in 10.0 g of Ar?

Chemistry

1 answer:

NARA [144]3 years ago

6 0

Answer:

1.51 x 10^23 atoms of Ar

Explanation:

Start out with what you're given.

10.0 g of Ar/1 * 1 mol/39.948 g Ar * 6.022*10^23 atoms/1 mol

(It's easier if you write it out.)

You always want to start out w/ what you're given and continue your unit conversions from there. Since you started out with grams of Argon, you want grams of Argon on the bottom to cancel it out. Once you found the unit conversion: 1 mol/39.948 g Ar (molar mass of Ar), now it's time to convert to atoms. Since 1 mol is up on top, put 1 mol on the bottom. Considering you're looking for atoms, use Avagradro's number: 6.022*10^23.

Once simplified, you should get 1.51 x 10^23 atoms of Ar.

***I hope this helps! I also suggest using a mole map, if you're interested.***

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Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol

erica [24]

Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

$2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O$

Molecules of ethyne = 52

Molecules of oxygen gas = 150

According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

$\frac{5}{2}\times 52=130 molecules$ of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

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2 years ago

Many grams of aluminum are required to produce 3.5 moles Al2O3 in the presence of excess O2?

Aleks [24]

The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below

write the equation for reaction
4 Al + 3O2 =2 Al2O3

by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles

mass of Al = moles / x molar mass

= 7 moles x27 g/mol =189 grams

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2 years ago

According to the concept of the tragedy of the commons, what happens to shared resources over time?

Maslowich

I would believe the answer to this question is D. According to the concept of the tragedy of the commons, shared resources are used by more than one organism. Due to the large consumption of shared resources they start to be fewer and fewer in number and over time if we are not careful they will be depleted.

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2 years ago

Read 2 more answers

Use the following equation to answer question 1-4. Make sure you balance first.

worty [1.4K]

<h3>Answer:</h3>

5.2 mol H₂O

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O

[Given] 10.4 mol HCl

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

[RxN] 6 mol HCl = 3 mol H₂O

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 10.4 \ mol \ HCl(\frac{3 \ mol \ H_2O}{6 \ mol \ HCl})$
Multiply/Divide: $\displaystyle 5.2 \ mol \ H_2O$

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2 years ago

How many moles of I2 will form 3.58 g of NI3?

Mila [183]

1. The balanced chemical reaction is:

N2 +3 I2 = 2NI3

We are given the amount of product formed. This will be the starting point of our calculations.

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Thus, 0.014 mol of I2 is needed to form the given amount of NI3.

6 0

3 years ago