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3 years ago

How are clouds formed? Simple answer please!

Chemistry

2 answers:

GaryK [48]3 years ago

7 0

Answer:

Clouds form when the invisible water vapor in the air condenses into visible water droplets or ice crystals. For this to happen, the parcel of air must be saturated, i.e. unable to hold all the water it contains in vapor form, so it starts to condense into a liquid or solid form.

Explanation:

MARK ME BRAINLIEST!!!!!!!! THANKS BESTFRIENNNNNDDDDD!!!!!!!!

Arada [10]3 years ago

6 0

Answer:

Explanation:

<em>I HOPE THIS HELPS*--- </em>*meow*

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Define the term<br>molecule?

professor190 [17]

Answer:

A termolecular reaction requires the collision of three particles at the same place and time.

Explanation:

This type of reaction is very uncommon because all three reactants must simultaneously collide with each other, with sufficient energy and correct orientation, to produce a reaction.

7 0

3 years ago

In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera

vredina [299]

Answer : The enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $1.55J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 107.70 g

$\Delta T$ = change in temperature = $T_2-T_1=(22.80-20.34)=2.46^oC$

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

$q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]$

$q=1112.3J=1.1123kJ$

Now we have to calculate the enthalpy change of dissolution of $KClO_4$

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of $KClO_4$ = 2.84 g

Molar mass of $KClO_4$ = 138.55 g/mol

$\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole$

$\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole$

Therefore, the enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

5 0

4 years ago

Xco3 enter the group number of x

Mademuasel [1]

We know that C has a formal charge of +4 while each O element has a formal charge of -2, therefore the formal charge of X is:

X + 2 – 2 * 3 = 0

X + 2 – 6 = 0

X = 4

<span>So X has a group number of 4.</span>

4 0

3 years ago

Consider the reaction for the dissolution of solid magnesium hydroxide.

sp2606 [1]

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

<em>Where X is solubility.</em>

<em />

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

<h3>Molar solubility is 1.12x10⁻⁴M</h3>

4 0

3 years ago

How would you solve the following geometry problem?: ∆PQR, find the measure of ∡P. In Triangle PQR where angle Q is a right angl

harina [27]

Your solution steps look correct, but the answer was not completed. Given that QR = 33.8 and PQ = 57.6, and Q is the right angle, then tan P = opposite / adjacent = QR / PQ = 33.8 / 57.6 = 0.586.
Then tan P = 0.586
Angle P = tan^-1 (0.586) = 30.4 degrees.

7 0

4 years ago