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Katyanochek1 [597]
3 years ago
10

If you start with 30 moles of O2, how many moles of SiO2 can you make?

Chemistry
1 answer:
Ivanshal [37]3 years ago
7 0

Answer:

30moles of SiO₂

Explanation:

Given parameters:

Number of moles of O₂  = 30moles

Unknown:

Number of moles of SiO₂  = ?

Solution:

To solve this problem, we need to write the reaction expression:

       Si +  O₂  →  SiO₂  

The reaction is balanced;

      1 mole of O₂ will produce 1 mole of SiO₂

     30mole of O₂ will produce 30moles of SiO₂

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Given the following equation, how many grams of PbCO3 will dissolve when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3?
9966 [12]
Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+

From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3

Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3
4 0
3 years ago
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Mass = 200.0 g volume = 100.0cm3 What is the density?
Marianna [84]
The density of an object is defined as its mass divided by its volume. Mathematically, density = Mass / Volume. The unit of density is kilogram per cubic meter, kg / m^3 or g /cm^3.
For the question given above: the 
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Volume = 100.0 cm^3
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3 0
3 years ago
1) How many moles are in 4.0x10^24 atoms?
-Dominant- [34]

Answer:

<h2>6.64 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{4 \times  {10}^{24} }{6.02 \times  {10}^{23} }   \\  = 6.644518...

We have the final answer as

<h3>6.64 moles</h3>

Hope this helps you

3 0
3 years ago
I will mark brainliest
Veronika [31]

I believe the answer is B??????????? Hope this helps

~Queensupreme

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a_sh-v [17]

Answer:

Explanation:

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3 years ago
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