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Solnce55 [7]
3 years ago
10

This gas law takes into consideration temperature and pressure

Chemistry
2 answers:
grandymaker [24]3 years ago
5 0

Answer:

A. charles law

lisov135 [29]3 years ago
3 0
Yeah yeah the answer should be A
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1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
Consider the following intermediate chemical equations.
vfiekz [6]

Answer:

-250.3kJ

Explanation:

Based in the reactions and using -<em>Hess's law-</em>:

(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ

(2) 4PCl₅(g) → P₄(s) + 10Cl₂  ΔH₂ = 3438kJ

The sum of (1) + (2) is:

4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ

Dividing this reaction in 4:

PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = <em>-250.3kJ</em>

8 0
3 years ago
Heat is being transferred from the ground to the air above it by conduction. Explain how this is possible.
Klio2033 [76]

Answer:

Conduction directly affects air temperature only a few centimeters into the atmosphere. During the day, sunlight heats the ground, which in turn heats the air directly above it via conduction. At night, the ground cools and the heat flows from the warmer air directly above to the cooler ground via conduction.

Explanation:

4 0
3 years ago
Read 2 more answers
PLEASE HELP WITH ALL
Gala2k [10]

Answer:

Question 1: D

Question 2: B C E F

Question 3: A

Question 4: A

Question 5: A

5 0
3 years ago
Read 2 more answers
Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answ
inessss [21]

Answer:

\Delta S^{0} for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: \frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)

\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]

where S^{0} represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K

So, \Delta S^{0} for the given reaction is -99.4 J/K

7 0
3 years ago
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