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Amiraneli [1.4K]
3 years ago
9

Chloroform (CHCl3), an important solvent, is produced by a reaction between methane and chlorine. CH4(g) + 3 Cl2(g) CHCl3(g) + 3

HCl(g) How many grams of CH4 is needed to produce 37.5 g CHCl3?
Chemistry
1 answer:
PIT_PIT [208]3 years ago
4 0
CH₄(g) + 3 Cl₂(g) → CHCl₃(g) + 3 HCl(g)
From the equation we notice that 1 mole of methane produces 1 mole of chloroform:
16 g Methane → 119.38 g Chloroform
?   g Methane → 37.5 g Chloroform
by cross multiplication:
= (16 * 37.5) / 119.38 = 5.0 g methane
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You have a raw egg you put it in a hot pan boom cooked egg most chemical changes are nonreversibal <span />
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Name an element in the third period (row) of the periodic table with a total of 4 3p electrons
Wittaler [7]

Answer:

sulfur

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sulfur has 4p electrons.

phosphorus has 3p electrons.

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3 0
3 years ago
How much heat is released when 15.7g of methane (c2h6) is combusted if the enthalpy of the reaction is - 1560.7 kj
lyudmila [28]

- 407.4 kJ of heat is released.

<u>Explanation:</u>

We have to write the balanced equation as,

2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O

Here 2 moles of ethane reacts in this reaction.

Now we have to find out the amount of ethane reacted using its given mass and molar mass as,

2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆

Heat released = ΔH × given mass / 60.14

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5 0
3 years ago
Using VSEPR theory which of the following would be the correct shape for nitrogen trifluoride?
RUDIKE [14]

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trigonal pyramidal

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In NF3, the nitrogen atom is sp3 hybridized. Now we must remember that according to the VSEPR theory, the number of electron pairs in the valence shell of the central atom in a molecule determines its shape.

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However, due to the lone pair, the three fluorine atoms are arranged in a  trigonal pyramidal geometry. Hence the correct shape of the molecule is trigonal pyramidal.

6 0
3 years ago
Volume of HCl used 25.0mL 4 l
borishaifa [10]

Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB =  25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

8 0
3 years ago
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