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creativ13 [48]
3 years ago
8

What weather does spikes or icicles bring?

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
4 0
When it drops below 0 / freezing
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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
When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of
taurus [48]

Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where  ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C  and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c  or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

3 0
3 years ago
Question 34 (1 point)
Grace [21]

Answer:

8.33 atm

Explanation:

Xe is   5  out of (4+5)   or   5 / 9 ths of the gas present

  5/9  * 15 atm = 8.33 atm

4 0
2 years ago
What is another name for the acid dissociation constant?
sveticcg [70]
A. Ionization constant
4 0
3 years ago
Read 2 more answers
2.
Arturiano [62]

Answer:

they are producers and they have means to keep themselves warm

Explanation:

I took test and saw answer.

6 0
3 years ago
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