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Andrews [41]
3 years ago
7

Isotopes (such as hydrogen-1, hydrogen-2, and hydrogen-3) are atoms of the same element that differ in:

Chemistry
1 answer:
maria [59]3 years ago
8 0
They have a different amount of neutrons.
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The air pressure inside a submarine is 1.24 atm. What would be the height of a column of mercury (mm Hg) balanced by this pressu
Ipatiy [6.2K]
Since one atmosphere is 29.9213 inches of mercury, we can simply find this by taking 29.9213 and multiplying that by 1.24. You end up with 37.102412 inches of mercury. I hope this helps you out, and good luck on the rest of your homework.
6 0
3 years ago
C2h6 what is the name
My name is Ann [436]
Ethane
see the picture for explanation

8 0
3 years ago
How many grams of water are produced when oxygen reacts with an excessive amount of propane (C3H8)?
lora16 [44]

Answer:

From the equation you will see that 1 mol of propane generates 4 mols of water.

Since the molar mass of water  

M ( H2O)=2×1+16=18g/mol

2 mol propane will generate  

2 ×4×18=144g   of water

Explanation:

Since the molar mass of water  

M ( H2O)=2×1+16=18g/mol

2 mol propane will generate  

2 ×4×18=144g   of water

5 0
2 years ago
Which type of property can be observed only by changing a substance into a different substance
Nana76 [90]

Chemical Property

<u>Explanation:</u>

When a substance is changed into another substance, may be due to addition of some other substance or removal of water from that substance.

* its identity changes

*its reactivity changes

*reactivity towards water, air and many other substances will change

*its chemical composition changes

and all these changes occurs due to the chemical properties.

If the physical property changes then there will be change in the state of matter, but its boiling point, melting point , reactivity towards water, air and other substances will not change, and it will not lose its identity.

3 0
3 years ago
77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a
BigorU [14]

Answer:

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

Mass of water = 225 g

Initial Temperature of water = 22°C

Mass of Aluminium cup = 44 g

Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C

Final temperature of the setup = 26°C

Note that, specific heat capacity of water = 4.186 J/g°C

Let the specific heat of the unknown metal be C

Heat lost from the unknown metal

= (77)(C)(99 - 26) = (5,621C) J

Heat gained by water

= (225)(4.186)(26 - 22) = 3,767.4 J

Heat gained by Aluminium cup

= (44)(0.92048)(26 - 22) = 162.00448 J

Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

Hope this Helps!!!

6 0
3 years ago
Read 2 more answers
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