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vampirchik [111]
3 years ago
5

I really like chemistry but no barely anything about it so could you tell me something about it please

Chemistry
2 answers:
fomenos3 years ago
4 0

Answer:

2 hydrogen atoms and one oxygen atoms make h2o/water

VLD [36.1K]3 years ago
3 0
U could use khan academy to learn more ab it it’s basically like a free class
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A secondary step in the process to produce ultra-pure silicon is to combine silicon tetrachloride with magnesium. How many grams
mariarad [96]

The mass of silicon, Si produced from the reaction is 329.41 g

<h3>Balanced equation </h3>

SiCl₄ + 2Mg —> 2MgCl₂ + Si

Molar mass of SiCl₄ = 28 + (35.5×4) = 170 g/mol

Mass of SiCl₄ from the balanced equation = 1 × 170 = 170 g

Molar mass of Si = 28 g/mol

Mass of Si from the balanced equation = 1 × 28 = 28 g

From the balanced equation above,

170 g of SiCl₄ reacted to produce 28 g of Si.

<h3>How to determine the mass of Si produced </h3>

From the balanced equation above,

170 g of SiCl₄ reacted to produce 28 g of Si.

Therefore,

2 Kg (i.e 2000 g) of SiCl₄ will react to produce = (2000 × 28) / 170 = 329.41 g of Si

Thus, 329.41 g of Si were obtained from the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

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3 years ago
0.1005 liters is the same as cm cubed mL
Xelga [282]

Answer: 100.5cm³ and  100.5 mL

#answerwithquality #BAL

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3 years ago
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mote1985 [20]

Answer: C

Explanation:

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How many grams of hno3 would you need to prepare 5.5 l of this solution
k0ka [10]

8305 grams of HNO3 would be needed to prepare 5.5L of a solution. Details on how to calculate mass is found below.

<h3>How to calculate mass?</h3>

The mass of a substance in a solution can be calculated using the following formula:

Density = mass ÷ volume

According to this question, 5.5L of a HNO3 solution is given.

Density of HNO3 is 1.51 g/cm³

Volume of HNO3 = 5500mL

1.51 = mass/5500

mass = 8305g

Therefore, 8305 grams of HNO3 would be needed to prepare 5.5L of a solution.

Learn more about mass at: brainly.com/question/19694949

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2 years ago
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