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Zarrin [17]
3 years ago
15

Which glacial landforms require more than one glacier to form

Chemistry
2 answers:
ruslelena [56]3 years ago
7 0

Answer:

C. horns, arêtes, drumlins, and hanging valleys

grandymaker [24]3 years ago
5 0
<span>aretes,hanging valleys,horns,and drumlins here you go!</span>
You might be interested in
Ammonia, NH3; ammonium nitrate, NH4NO3; and ammonium hydrogen phosphate, (NH4)2HPO4, are all common fertilizers. Rank the compou
ValentinkaMS [17]

Answer:

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

Explanation:

Mass percentage -

Mass percentage of A is given as , the mass of the substance A by mass of the total solution multiplied by 100.

i.e.

mass % A = mass of A / mass of solution * 100

Given,  

mass of nitrogen = 14 g/mol

mass of hydrogen = 1 g/mol

mass of oxygen = 16 g/mol

mass of phosphorus = 31 g/mol

1. Ammonia, NH₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 1 * 14 g/mol  + 3 * 1 g/mol  

mass of solution = 17 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 17 g/mol * 100

mass % N = 82.35 % .

2.   ammonium nitrate, NH₄NO₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 4 * 1 g/mol  + 3 * 16 g/mol

mass of solution = 80 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 80 g/mol * 100

mass % N = 17.50 % .

3.   ammonium hydrogen phosphate, (NH₄)₂HPO₄

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

mass of solution = 2 * 14 g/mol  + 9 * 1 g/mol  + 4 * 16 g/mol + 1 * 31 g/mol

mass of solution = 132 g/mol

and ,

mass of nitrogen = 14 g/mol

The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 132 g/mol * 100

mass % N = 10.60 % .

Hence , the correct order is -

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

3 0
3 years ago
A student reacted NiS2 (MM = 122.83 g/mol) with O2 (MM=32.00 g/mol) to make SO2 (MM=64.07 g/mol) according to this balanced equa
Ilya [14]

Answer:

Percentage yield = 0.49 × 10² %

Explanation:

Given data:

Actual yield of SO₂ = 4.309 ×10² g

Theoretical yield of SO₂ = 8.78 ×10² g

Percentage yield = ?

Solution:

Chemical equation:

2NiS₂ + 5O₂ → 2NiO + 4SO₂

Percentage yield:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 4.309 ×10² g  / 8.78 ×10² g  × 100

Percentage yield = 0.491 × 100

Percentage yield = 49.1%

In scientific notation:

Percentage yield = 0.49 × 10² %

5 0
3 years ago
The octet rule states that, in chemical compounds, atoms tend to have _____.
olga nikolaevna [1]

The correct answer is C


4 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

<u>Thus 52.8 g of CO2 is produced.</u>

5 0
3 years ago
I need help fast plzzzzzzzzz
sineoko [7]
What’s the question?
6 0
3 years ago
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