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Mamont248 [21]
3 years ago
13

Which of the following is the best example of chemical weathering?

Chemistry
1 answer:
galina1969 [7]3 years ago
8 0
I believe the answer is B.

Hope this helps
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If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the re
lord [1]

Answer;

=259 ml

Explanation;

-According to Gay Lussac's Law of Combining Volumes when gases react, they do so in volumes which have a simple ratio to one another, and to the volume of the product formed if gaseous, provided the temperature and pressure remain constant.

-Thus; from the volume of nitrogen and oxygen gases; we have; 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas.

-Therefore, nitrogen gas is the limiting reactant, and for each mole of nitrogen gas used, we will get 1 mole of N2O. This means the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, thus, 100% yield will produce 316 mL.

However, with 82% yield the volume would be; 316 × 82/100 =259 ml

Therefore; the volume of N2O at 82% yield will be 259 ml

7 0
3 years ago
Read 2 more answers
A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man
max2010maxim [7]

Answer:

There will react 0.400 moles of oxygen.

Explanation:

<u>Step 1:</u> Data given

Volume of the closed flask = 200.00 mL = 0.2 L

Number of moles of CO = 2.000 mol

Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

<u>Step 2</u>: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

<u>Step 3</u>: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

<u>Step 4:</u> When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

<u>Step 5:</u> Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

⇒ V = the volume = 200.0 mL = 0.2 L

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

n = 3.6 moles = total number of moles

<u>Step 6:</u> Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

Moles O2 = (2-X) moles

Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

6 0
3 years ago
14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and d
zepelin [54]

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock solution.

M_2\text{ and }V_2 are the molarity and volume of diluted solution.

From data (A) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

5 0
3 years ago
How are atoms arranged in molecular compounds?
frutty [35]
Atoms<span> are </span>arranged in molecular compounds in groups. 
<span>
For </span>covalent compounds<span>: </span>
<span>consider drawing the lewis structure of the covalent compound in question, putting the atom which is least electronegative (save hydrogen) in the center.


</span>
4 0
3 years ago
Para que sirven las rondas??<br> ayuda por favor
prisoha [69]

Answer:

las rondas sirven para ayudar con problemas matematicos asi como cilindros

6 0
3 years ago
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