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Alexxandr [17]
2 years ago
11

Mention the significance of dams and irrigation canals? ​

Chemistry
1 answer:
GarryVolchara [31]2 years ago
3 0

<u>Significance</u><u> </u><u>of</u><u> </u><u>dams</u><u> </u><u>and</u><u> </u><u>canals</u><u> </u><u>are</u><u> </u><u>:</u><u>-</u><u> </u>

  • Water is stored in dams that are built across rivers. The water is then supplied to nearby towns and cities through pipelines

  • Water for agricultural purposes is supplied through a system of irrigation canals

<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>~</em>

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Which of the following gas samples would contain the same amount of gas as 200 mL of helium, He(g), at 25° C and 1 atm?
monitta

Answer:

200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm (the second choice) would contain an equal number of gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\! (assuming that all four gas samples behave like ideal gases.)

Explanation:

By Avogadro's Law, if the temperature and pressure of two ideal gases is the same, the number of gas particles in each gas would be proportional to the volume of that gas.

All four gas samples in this question share the same temperature and pressure. Hence, if all these gases are ideal gases, the number of gas particles in each sample would be proportional to the volume of that sample. Two of these samples would contain the same number of gas particles if and only if the volume of the two samples is equal to one another.

The second choice, 200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm, is the only choice where the volume of the sample is also 200\; \rm mL \!. Hence, that choice would be the only one with as many gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\!.

7 0
3 years ago
Different between:Electrovalent bond and covalent bond​
nikdorinn [45]

Explanation:

electrovalent bond covers single bond.

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Gold forms face-centered cubic crystals. The atomic radius of a gold atom is 144 pm. Consider the face of a unit cell with the n
kipiarov [429]

Answer:

The length of an edge of this unit cell is 407.294 pm

Explanation:

Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.

To determine the edge length, a relationship between the radius of the atom and edge length is used.

X = R√8

Where;

X is the length of an edge of this unit cell

R is the radius of the gold atom = 144 pm = 144 X 10⁻¹² m

X = 144 X 10⁻¹²√8

X = 407.294 X 10⁻¹² m

X = 407.294 pm

Therefore, the length of an edge of this unit cell is 407.294 pm

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<span>C + O2 → CO2 (8,376,726 tons) x (0.80) / (12.01078 g C/mol) x (1 mol CO2/ 1 mol C) x (44.00964 g CO2/mol) = 24,555,054 tons CO2</span>
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