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4 years ago

How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?

Chemistry

1 answer:

storchak [24]4 years ago

7 0

Given:

2.5 mol O2

Required:

Grams of water

Solution:

2H2 + O2 -> 2H2O

Moles of H2O = (2.5 mol O2)(2 moles H2O/1 mole O2) = 5 moles H2O

Molar mass of H2O = 18.02 g/mol

Mass of H2O = (5 moles H2O)( 18.02 g/mol)

Mass of H2O =90.1g H2O

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The temperature of a 350. mL sample of gas increases from 27 °C to 227 °C. What is the final volume of the sample of gas, if the

Liono4ka [1.6K]

Answer:

$V_2=583.3mL$

Explanation:

Hello,

In this case, for the given constant amount and pressure of the, we apply the Charles' law which allows us to understand the volume-pressure behavior as a directly proportional relationship:

$\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Thus, since we need to compute the volume after the temperature increase (which must be expressed in absolute Kelvins), we obtain V2 as:

$V_2=\frac{V_1T_2}{T_1} =\frac{350.0mL*(227+273.15)K}{(27+273.15)K}\\ \\V_2=583.3mL$

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3 years ago

How to describe the rotation and revolution of the moon??

Alenkasestr [34]

The moon completes one rotation and one revolution every month.

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4 years ago

Based on graph 4 across a period the electronegativity

Romashka-Z-Leto [24]

4

Answer with explanation

7 0

3 years ago

Read 2 more answers

The hydrolysis of esters in base is called ________. a. the hunsdiecker reaction b. transesterification c. saponification d. the

KengaRu [80]

The hydrolysis of esters in base is called saponification .

So, option C is correct one.

The saponification is the process that involves conversion of fats , oils , lipids into soap and water in the presence of alkaline medium. Saponification is the process of making soap.

During the saponification process, the mixture has an acidity, which tells that it's not safe for usage. After the saponification process is complete, the pH should be a base.The process of formation of carboxylic salt and water by hydrolysis of ester in base is called saponification.

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7 0

2 years ago

1.00 mole of an ideal gas at STP is cooled to -41°C while the

qwelly [4]

Answer:

V₂ = 18.13 L

Explanation:

Given data:

Mole of gas = 1 mol

Initial temperature = 273 K

Initial pressure = 1 atm

Final volume = ?

Final temperature = -41°C (-41+273 = 232 K)

Final pressure = 805 mmHg (805/760 = 1.05 atm)

Solution:

First of all we will calculate the initial volume of gas.

PV = nRT

V = nRT/P

V = 1 mol × 0.0821 mol.L/atm.K × 273 K / 1 atm

V = 22.4 L/atm / 1 atm

V = 22.4 L ( initial volume)

Now we will determine the final volume by using equation,

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Now we will put the values.

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 1 atm × 22.4 L × 232 K / 273 K × 1.05 atm

V₂ = 5196.8 atm .L. K / 286.65 atm.K

V₂ = 18.13 L

5 0

3 years ago