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alisha [4.7K]
2 years ago
7

Use the scenario below to answer the following

Chemistry
1 answer:
Rom4ik [11]2 years ago
5 0

Answer:

i had this question a couple of weeks ago i just cant remember the answer other wise i would tell you but im sorry:( <3

Explanation:

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5.2 G of HCl are dissolved in 25 mL of solution. What is the molarity?
Kobotan [32]

Answer:

0.21 M

Explanation:

Molarity is the calculation of the solution in which the number of solute per liter of the solutions. It is the most common measurement unit that is used to measure the concentration of the solution.

The molarity is the unit that is used to measure or calculate the volume of the solvent. The amount of solvent is used in the chemical reaction.

The amount of the two solvent in the same quantity is measured by the formula called c1v1 and c2v2.

6 0
3 years ago
An objects volume can be found by adding its mass by its?
Inessa05 [86]
Volume* density=MASS
4 0
3 years ago
The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol.
alexira [117]
<span>the balanced chemical equation for the reaction is as follows;
 C</span>₃H₈ + 5O₂ ---> 3CO₂ + 4H₂<span>O
                   
stoichiometry of </span> C₃H₈ to O₂ is 1:5
   
number of moles of  C₃H₈ reacted - 0.025 g / 44.1 g/mol = 0.000567 mol according to molar ratio of 1:5
number of O₂ moles required are 5 times the amount of  C₃H₈ moles reacted therefore number of O₂ moles required - 0.000567 x 5 = 0.00284 mol .
mass of O₂ required - 0.00284 mol x 32.00 g/mol = 0.091 mol .
 answer is 0.091 mol
5 0
3 years ago
Read 2 more answers
Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What
Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For CH_4  :-

Mass of CH_4  = 1.28 g

Molar mass of CH_4  = 16.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.28\ g}{16.04\ g/mol}

Moles_{CH_4}= 0.0798\ mol

For O_2  :-

Mass of O_2  = 10.1 g

Molar mass of O_2  = 31.998 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{10.1\ g}{31.998\ g/mol}

Moles_{O_2}= 0.3156\ mol

According to the given reaction:

CH_4+2O_2\rightarrow CO_2+2H_2O

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

<u>Limiting reagent is the one which is present in small amount. Thus, CH_4 is limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0798\ moles= \frac{Mass}{44.01\ g/mol}

Mass of CO_2 = 3.51 g

<u> Theoretical yield = 3.51 g</u>

3 0
3 years ago
I need filling the blank spots ​
AnnZ [28]

Answer:

have you tried c

Explanation:

the chicken and I don't know if you can make it

4 0
2 years ago
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