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3 years ago

Use the balanced equation to answer each question.

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

5 0

Answer:

2 moles of C2H6 are required to form 4 moles of CO

therefore for 2 moles of CO, 1 mole of C2H6 is required

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4 years ago

What is the penetrating power of beta

zlopas [31]

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3 years ago

The data showed that recently the alligator population has decreased. How could the decrease in the alligator

Marina CMI [18]

Answer:

Since the alligator population decreased, this means there are less predators for the gar. Less predators for the gar mean the gar population will increase. More gar also means more competition for the diving duck in terms of animals that eat crabs, and thus the diving duck population will also decrease.

Explanation:

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3 years ago

Kemmi pipets 25.00 ml of pure 1-propanol (c3h7oh, a liquid organic alcohol) into a 100.0 ml volumetric flask. she dilutes it wit

algol13

Q1)
As Kemmi pipettes a volume of 25.00 ml of the solution
density of pure propanol is 0.803 g/ml
This means that in 1000 ml of solution - 0.803 g of pure propanol
Therefore in 25.00 ml of solution - 0.803 g x 25.00 ml / 1000 ml
= 0.0201 g
Using molar mass, number of moles can be calculated= 0.0201 g / 60.09 g/mol
= 3.35 x 10⁻⁴ mol
therefore the number of pure propanol moles in exactly 25.00 ml is
3.35 x 10⁻⁴ mol

Q2)
molarity is the concentration of the solution. It can be defined as the number of moles of solute per liter of solution
we know the number of moles in 25.00 ml of solution. When its diluted in a 100.00 ml volumetric flask, number of moles remain constant but now the volume over which the moles of solute are dissolved is increased.
therefore number of moles = 3.35 x 10^(-4) mol
volume over which its dissolved - 100.00 / 10³ dm³
= 1.0000 x10⁻¹ dm³
the molarity = 3.35 x 10⁻⁴ mol / 1.0000 x10⁻¹ dm³
= 3.35 x 10⁻³ mol/dm³

5 0

4 years ago

Aluminum nitrate reacts with sodium hydroxide to form a precipitate of aluminum hydroxide. What mass of aluminum hydroxide will

lora16 [44]

Answer:

The mass of aluminum hydroxide that will precipitate when 35.5 mL of 0.145 mol/L aqueous aluminum nitrate is mixed with 44.2 mL of 0.215 mol/L aqueous sodium hydroxide is approximately 0.247 grams

Explanation:

The given parameters are;

The volume of 0.145 mol/L aqueous aluminum nitrate in the reaction = 35.5 mL

The volume of 0.215 mol/L aqueous sodium hydroxide in the reaction = 44.2 mL

The balanced chemical equation for the reaction is given as follows;

Al(NO₃)₃ (aq) + 3NaOH (aq) → Al(OH)₃ (s) + 3NaNO₃ (aq)

Therefore;

1 mole of Al(NO₃)₃, reacts with 3 moles of NaOH to form 1 mole Al(OH)₃, and 3 moles of NaNO₃

The number of moles of aqueous aluminum nitrate in the reaction = 35.5/1000 L × 0.145 mol/L = 0.0051475 moles

The number of moles of aqueous sodium hydroxide in the reaction = 44.2/1000 L × 0.215 mol/L = 0.009503 moles

Therefore, 0.009503 moles NaOH will react with 0.009503/3 moles Al(NO₃)₃ to form 0.009503/3 moles of aluminum hydroxide precipitate

The molar mass of aluminum hydroxide = 78 g/mol

Therefore, 0.009503/3 moles of aluminum hydroxide weighs 78 × 0.009503/3 = 0.247078 grams

The mass of aluminum hydroxide that will precipitate out ≈ 0.247 grams.

7 0

3 years ago