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3 years ago

Which elements are included in group 5a

Chemistry

1 answer:

RUDIKE [14]3 years ago

7 0

Answer:

Group 5A (or VA) of the periodic table are the pnictogens:

the nonmetals nitrogen (N),

and phosphorus (P),

the metalloids arsenic (As)

antimony (Sb),

the metal bismuth (Bi).

and moscovium (Mc).

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The carbon dioxide exhaled by astronauts can be removed from a spacecraft by reacting it with lithium hydroxide (LiOH). The reac

maks197457 [2]

Two astronauts would exhale about 40 moles of carbon dioxide daily.

Carbon dioxide reacts with lithium hydroxide in a 1 : 2 mole ratio. Set up a proportion:

1 : 2 = 40 : x

Then, find x: 12=40x

 Cross multiply. x = 80 moles of LiOH per day for both astronauts

8 0

4 years ago

Which reacts faster, rock salt or grains of salt? Why?

sweet-ann [11.9K]

Answer:

salt

Explanation:

because it has a much larger volume ratio

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3 years ago

Oxidation number is

zavuch27 [327]

The correct letter to your answer would be A

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3 years ago

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = 181,66 kJ/mol

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = 0,216 kJ/Kmol

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ 4. +117,1 kJ/mol

I hope it helps!

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3 years ago

The enthalpy change for converting 1.00 mol of ice at -50.0°c to water at 70.0°c is __________ kj. the specific heats of ice, wa

kogti [31]

First, calculate for the amount of heat used up for increasing the temperature of ice.

H = mcpdT
H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J

Then, solve for the heat needed to convert the phase of water.
H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J

Then, solve for the heat needed to increase again the temperature of water.
H = (18 g)(4.18 J/gK)(70 k)
H = 5266.8 J

The total value is equal to 13157.8 J

Answer: 13157.8 J

8 0

3 years ago