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4vir4ik [10]
3 years ago
8

Why is helium considered as a noble gas

Chemistry
2 answers:
dolphi86 [110]3 years ago
7 0
Noble gases are known for having a full outer shell of electrons which helium has as it has two electrons its first electron shell is completely filled

larisa [96]3 years ago
3 0
Helium is  considered a noble gas because it has 8 valence electrons, as well as it also is a stable element. Meaning that it consists of a full octet. 
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Thunderstorms between Tampa and titusville
3 0
3 years ago
Read 2 more answers
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
2 years ago
A 6.0 L container of gas is 2.5atm. What would be the pressure if the volume is 12.0L
Scorpion4ik [409]
This works because it demonstrates that as volume increases, pressure decreases (inverse relationship)

4 0
2 years ago
Given the following values for the heats of formation, what is the number of moles of ethane (C2H6, MW 30.0) required to produce
baherus [9]

Answer:

0.641 moles of ethane

Explanation:

Based on the equation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

We can determine ΔH of reaction using Hess's law. For this equation:

<em>Hess's law: ΔH products - ΔH reactants</em>

ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}

<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>

<em />

ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}

ΔH = -1559.7kJ/mol

That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:

1.00x10³kJ * (1mole ethane / 1559.7kJ) =

<h3>0.641 moles of ethane</h3>
7 0
2 years ago
If the half life is 6.3 minutes how much of a 128.0 milligram sample will remain after 15 minutes​
zaharov [31]

Answer:

24.5

Explanation:

15/6.3 => 2.38 half lives passed.

.5^2.38 => 0.19198 decimal representation of the percentage that is left over after 2.38 half lives have passed.

0.19198 *128 = 24.5 mg of the material remaining.

6 0
2 years ago
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