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3 years ago

Which ocean surrounds Antarctica at the South Pole?

Chemistry

2 answers:

tatuchka [14]3 years ago

7 0

the answer on edge is SOUTHERN OCEAN on edge

Sergeeva-Olga [200]3 years ago

4 0

Answer:

Southern Ocean

Explanation:

The Southern ocean is also referred to as the Antarctic ocean. It is the oceans below the latitude 60 degrees South. The currents in this oceans (Antarctic Circumpolar Current) circulate around Antarctica also shielding it from warmer waters from the equatorial regions. This keeps the Antartic region cold.

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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

3 years ago

Which of the following showing the Law of Conservation of Mass?

Stells [14]

The only one I know for sure is Mass is always conserved In a Chemical reaction.

7 0

3 years ago

Read 2 more answers

Rock and sand are non-living.all organisms are ____

Sladkaya [172]

Rock and sand are non-living. All organisms are living.

7 0

3 years ago

Read 2 more answers

3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O

AysviL [449]

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 3 moles
HNO₃: 8 moles
Cu(NO₃)₂: 3 mole
NO: 2 moles
H₂O: 4 moles

The molar mass of the compounds is:

Cu: 63.55 g/mole
HNO₃: 63 g/mole
Cu(NO₃)₂: 187.55 g/mole
NO: 30 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Cu: 3 moles ×63.55 g/mole= 190.65 grams
HNO₃: 8 moles ×63 g/mole= 504 grams
Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
NO: 2 moles ×30 g/mole= 60 grams
H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

$mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3} x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }$

<u><em>mass of Cu(NO₃)₂= 5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

Learn more about the reaction stoichiometry:

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7 0

2 years ago

Summarize your findings in a short report of 150 words. Include your hypothesis, observations, data, calculations, and conclusio

Damm [24]

Answer:

Follow these steps.

1. Fill the matchbox with pebbles. Weigh the matchbox with the pebbles inside. Record that weight.

2. Tie the string to the box. Allow the string to hang over the edge of the table.

3. Tie the other end of the string to a corner of the plastic bag, leaving an opening to put in coins.

4. Add coins one by one until the box is pulled off the table.

5. Count and record the number of coins and the weight of the bag with the coins in it.

6. Lay the round sticks on the table about 1 inch apart and about 2 inches from the edge of the table.

7. Put the matchbox on the rollers farthest from the edge of the table.

8. Now add coins one by one to the bag until the box is pulled off the table.

9. Count and record the number of coins and the weight of the bag with the coins in it.

10. Repeat the experiment. Determine your margin of error if your results vary. For accuracy, repeat the experiment if desired.

11. Using the equation for the coefficient of friction in the text above, determine the coefficient of friction for the matchbox in each experiment. Include this data in your summary.

Explanation:

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5 0

2 years ago