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Fudgin [204]
2 years ago
6

Energy removal is illustrated in. A. Changing water ice to waterB. Changing water to steamC. Boiling of gasolineD. Evaporation o

f sea water
Chemistry
1 answer:
Kryger [21]2 years ago
6 0

Answer:

D. Evaporation of sea water

Explanation:

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if an element [x] contains 8 protons, 8 neutrons and 8 electrons, the atomic number of the element is ?
Kaylis [27]

Answer:

8

Explanation:

The atomic number refers to the number which identifies the element, which is the proton number.

7 0
3 years ago
Ob] The electrodes of a fuel cellare in contact with water and all
iogann1982 [59]
Iron rusts when in contact with water.
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3 years ago
Use the equation n2 + 3h2=2nh3 if 8.0g N2 react, how many grams of NH3 will be produced
Eduardwww [97]

the balanced equation for the formation of ammonia is

N₂ + 3H₂ ---> 2NH₃

molar ratio of N₂ to NH₃ is 1:2

mass of N₂ reacted is 8.0 g

therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol

according to the molar ratio,

1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant

therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃

therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g

a mass of 9.72 mol of NH₃ is formed

6 0
3 years ago
Nyatakan maksud bagi kation​
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6 0
2 years ago
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A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the
alexdok [17]

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.4 atm

P_2 = final pressure of gas = 1.0 atm

V_1 = initial volume of gas = 250 mL

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 15^oC=273+15=288K

T_2 = final temperature of gas = 27^oC=273+27=300K

Now put all the given values in the above equation, we get:

\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}

V_2=625mL

Therefore, the volume of the bubble is, 625 mL

7 0
3 years ago
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