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Keith_Richards [23]
3 years ago
13

If HJ = 7x – 27, find the indicated values.

Chemistry
1 answer:
Sever21 [200]3 years ago
6 0
Lubos na paumanhin. maaari mo bang ipaliwanag muli ang tanong? sa totoo lang galing ako sa phillipines, masasabi mo ba sa fillipino; wika ko
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At constant temperature the pressure on a 6.0 L sample of a gas is reduced from 2.0 atm to 1.0 atm. What is the new volume of th
kotegsom [21]
P1V1=P2V2, so P1V1/P2=V2.
2atm x 6.0 L/1.0 atm = 12.0 L
The new volume would be 12.0 Liters
4 0
3 years ago
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Positrons are spontaneously emitted from the nuclei of
Leto [7]
Positrons are spontaneously emitted from the nuclei of potassium -37.
4 0
3 years ago
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Convert 99 moles of CO2 to atoms
lorasvet [3.4K]

Answer:

              1.78 × 10²⁶ Atoms

Explanation:

Relation between number of molecules and moles is,

No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol

No. of Molecules = 99 mol × 6.022 × 10²³ Molecules/mol

No. of Molecules = 5.96 × 10²⁵ Molecules

Also, In CO₂ Molecule there are 3 atoms.

So,

No. of atoms = 5.96 × 10²⁵ Molecules × 3

No. of atoms = 1.78 × 10²⁶ Atoms

3 0
2 years ago
Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> 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  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(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
When a cold front is approaching, what happens?
Marta_Voda [28]

Answer:There is colder and drier air coming

Explanation:

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2 years ago
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