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Kobotan [32]
3 years ago
15

Where does condensation occur?

Chemistry
1 answer:
Elanso [62]3 years ago
8 0

Answer:

A. Atmosphere

Explanation:

hope this helps

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SCIENCE!!!! PLEASE HELP!!! 20 POINTS
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As the Earth rotates, it also moves, or revolves, around the Sun. ... As the Earth orbits the Sun, the Moon orbits the Earth. The Moon's orbit lasts 27 1/2 days, but because the Earth keeps moving, it takes the Moon two extra days, 29 1/2, to come back to the same place in our sky.

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Give an example where handpicking is used for separation. (1)​
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If you want to separate black grapes from the mixture of black and green grapes, then you will simply pick black grapes using your hands from the mixture. In this way you are actually using handpicking separation method.

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Why would satellite imagery be more useful than a map in some instances? provides landmarks, such as buildings can be used when
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Write a balanced chemical equation showing how each metal oxide reacts with HCl. SrO Na2O Li2O BaO
Agata [3.3K]

Answer:

See below  

Explanation:

Each metal oxide reacts with HCl to form water and the metal chloride

\rm SrO + 2HCl \longrightarrow SrCl_{2} + H_{2}O\\\\Na_{2}O + 2HCl \longrightarrow 2NaCl + H_{2}O\\\\Li_{2}O + 2HCl \longrightarrow 2LiCl + H_{2}O\\\\BaO + 2HCl \longrightarrow BaCl_{2} + H_{2}O

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During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
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