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konstantin123 [22]
3 years ago
11

Three of the primary components of air are

Chemistry
1 answer:
Darya [45]3 years ago
8 0

Answer:

125.681 torr

Explanation:

The formula to calculate the partial pressure of oxygen:

P(O2) = Pcommon - PCO2 - PN2

1 atm = 760 torr

Therefore you need to find the partial pressure of oxygen:

P(O2) = 760 torr -0.285 torr - 634.034 torr =130.013 torr

- Hope that helped!

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For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)
juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = <u>4.82 grams benzene</u>

<u />

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

<u>Mass CO2 = 9.15 grams</u>

4 0
3 years ago
¿De qué manera puedo identificar ácidos y bases en la vida cotidiana?​
inna [77]

Answer:

Ya sea que explore su cocina, su baño o su cuarto de lavado, encontrará excelentes ejemplos de ácidos y bases a su alrededor. La escala de pH es un concepto importante tanto dentro como fuera del laboratorio de química.

4 0
3 years ago
why is it important different types of chemical reactions (i.e acid-base, precipitation, and oxidation-reduction reactions)?
stepladder [879]

Precipitation, acid-base, and oxidation-reduction reactions are the three most common types of chemical reactions.

When dissolved substances react, one (or more) solid products are produced, which is known as Precipitation reaction. These kinds of reactions, which are also occasionally known as double displacement, double replacement, or metathesis reactions, frequently involve the exchange of ions between ionic compounds in aqueous solutions. These interactions, which occur frequently in nature, are what cause kidney stones in mammals and coral reefs to grow in ocean waters. They are extensively utilized in industry to produce a variety of common and specialized compounds. Many chemical analysis procedures, such as gravimetric methods for figuring out the composition of matter and spot tests used to identify metal ions, heavily rely on precipitation reactions as well. This is called Precipitation reaction.

To learn more about Precipitation reaction visit here;

brainly.com/question/24158764

#SPJ4

5 0
1 year ago
What is the freezing point of a solution that contains 36.0 g of glucose in 500.0 g of water (Kf for water is 1.86C/m. The molar
lbvjy [14]
This freezing point business is usually based on molality, that is moles solute per kilogram of solvent. 

<span>molality = freezing point depression over Kf </span>

<span>In this case molality -10.3 degrees over -1.85 degrees Kf = 5.53 molal </span>

<span>This 5.53 molal solution is made up of l000 gms water and 5.53 moles glucose at 180 grams per mole for a total mass of 1997 grams </span>

<span>It volume would be l997 gms over 1.50 gms/ml or 1331 ml </span>

<span>We know that we have 5.53 moles of glucose dissolved in l331 ml of solution so now we can find how many moles of glucose in l000 ml or one liter of solution and this will be our Molarity </span>

<span>5.53 moles glucose over l331 ml = X moles glucose over l000 ml solution </span>

<span>cross multiply and solve for X moles glucose per liter solution </span>

<span>X = 4.15 moles glucose per liter = 4.15 Molar</span>
5 0
3 years ago
Convert 0.0338 moles of K3PO4 to grams.
Phantasy [73]
0.0338 K3PO4 212.27 g K3PO4
——————— x ————————-= 7.17g K3PO4
1 1 mol K3PO4

0.0338x 212.27= 7.174726

Answer : 7.17 g K3PO4
8 0
2 years ago
Read 2 more answers
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