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alexandr402 [8]
3 years ago
7

What is the volume occupied by 0.106 mol of helium gas at a pressure of 0.99 atm and a temperature of 309 K ?

Chemistry
1 answer:
iris [78.8K]3 years ago
6 0
0.99 x volume= 0.106 x 0.082 x 309
volume= 2.71 ml
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A type of friction that opposes the motion of an object traveling either through liquid or gas
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type of friction that opposes the motion of an object traveling either through liquid or gas is known as <u>drag</u><u>.</u>

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Who will win the super bowl? Kansas City Chiefs Or the Tampa Bay Buccaneers
Serga [27]

Answer: Kansas City Chiefs

Explanation: what ever you think.

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The boat shown in the photo below is moving along at a constant 20 miles per hour. Is the boat accelerating? Question 3 options:
SVETLANKA909090 [29]

Answer:

A

Explanation:

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4 0
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
128g of sulphur are burned in excess oxygen. What mass of sulphur dioxide forms? S + O2 --&gt; SO2 *
Rasek [7]

Mass of Sulphur dioxide : 256 g

<h3>Further explanation</h3>

Given

Reaction

S + O2 --> SO2 *

Required

Mass of Sulphur dioxide

Solution

mol of Sulphur (Ar=32 g/mol) :

mol = mass : Ar

mol = 128 : 32

mol = 4

From the equation, mol ratio S : SO2 = 1 : 1, so mol SO2 = 4

Mass of SO2 :

mass = mol x MW SO2

mass = 4 x 64

mass = 256 g

5 0
3 years ago
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