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oksian1 [2.3K]
3 years ago
8

0.0524 moles of H2O equals how many molecules of water?

Chemistry
2 answers:
Charra [1.4K]3 years ago
8 0
Hey there!:

<span>Use the Avogadro constant :
</span>
1 mole -------------- 6.02*10²³ molecules
0.0524 moles -------- molecules H2O

molecules H2O = 0.0524* ( 6.02*10²³) / 1

  molecules H2O = 3.15*10²² / 1

= 3.15*10²² molecules H2O




NeTakaya3 years ago
8 0

Answer:

Molec_{H_2O}=3.16x10^{22}MolecH_2O

Explanation:

Hello,

In this case, by applying the Avogradro's number as a relationship between moles and molecules, we obtain:

Molec_{H_2O}=0.0524molH_2O*\frac{6.022x10^{23}MolecH_2O}{1molH_2O} \\Molec_{H_2O}=3.16x10^{22}MolecH_2O

Best regards.

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3 years ago
Element in the noble ga family are considered stabled because their outer energy levels are
blsea [12.9K]

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Explanation: For noble gases they are stable in state since their outer shell contain fully occupied having 8 electrons.

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3 years ago
Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---&gt; CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ  

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

_________________________________

The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

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(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


7 0
3 years ago
PLEASE HELP!!!
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Answer:

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Explanation:

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The volume of the rock, V = 10 cm³

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d=\dfrac{m}{V}\\\\d=\dfrac{15\ g}{10\ cm^3}\\\\d=1.5\ g/cm^3

So, the density of the rock is equal to 1.5\ g/cm^3.

3 0
3 years ago
Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.
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Answer:

Answer :

The total pressure inside the cylinder is, 6.7 atm

The mole fraction of N_2 in the mixture is, 0.52

Solution :

First we have to calculate the total pressure inside the cylinder.

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

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Now put all the given values is expression, we get the total pressure inside the cylinder.

P_T=3.5+2.8+0.25+0.15=6.7atm

Now we have to calculate the mole fraction of N_2 in the mixture.

Formula used :

pN_2=XN_2 x P_T

where,

P_T = total pressure = 6.7 atm

pN_2 = partial pressure of nitrogen gas = 3.5 atm

XN_2 = mole fraction of nitrogen gas = ?

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

3.5atm=XN_2 x 6.7atm

XN_2=0.52

Therefore, the total pressure inside the cylinder is, 6.7 atm and the mole fraction of N_2 in the mixture is, 0.52

Hope it helps answer the question:)

5 0
3 years ago
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