Which element forms the skeleton of this polymer?

For each equation , write all possible mole ratios . A. 2HgO(s)>2Hg(l)+O2(g) B. 4NH3(g)+6NO(g)>5N2(g)+6H2O(l)

Dafna11 [192]

A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).

1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).

2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.

3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.

B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).

1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).

2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.

3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).

4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.

5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).

6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.

3 0

2 years ago

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How do the properties of the elements change as you move across a period on a periodic table from left to right?

Tomtit [17]

B) The elements become less reactive.

8 0

2 years ago

Which atom would it be most difficult to remove an electron from

Norma-Jean [14]

I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements). This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove. Examples Helium, Neon, Argon, Xenon, Krypton and Radon

6 0

2 years ago

PLEASE HELP ME WITH THIS ONE QUESTION!!

Ksenya-84 [330]

2 I think not 100% tho

5 0

2 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

2 years ago