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puteri [66]
2 years ago
7

Is oxygen more reactive than beryllium

Chemistry
2 answers:
laiz [17]2 years ago
8 0

Answer:

Yes oxygen is more reactive than beryllium

gladu [14]2 years ago
6 0

Elemental beryllium is very unreactive towards air and water, even at high temperatures.

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For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper
Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1\text{ and }V_1 are initial pressure and volume at initial temperature T_1.

P_2\text{ and }V_2 are final pressure and volume at initial temperature T_2.

We are given:

P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V

T_1=-25^oC=248.15 K

T _2 = 16^oC=289.15 K

Putting values in above equation, we get:

\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}

P_1=0.225 kPa

Hence, the new pressure will be 0.225  kPa.

4 0
3 years ago
What can be said about the spontaneity of this reaction? cdiamond→cgraphite
anastassius [24]
The gibbs free energy of the reaction of diamond to graphite is equal to -2.90 kJ/mol. The free energy is negative which means that the reaction is spontaneous. Therefore, the forward reaction is favored. Hope this helps. Have a nice day.
7 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
3 years ago
What is balance equation and how we balance equation
Elenna [48]

Answer:

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.

Explanation:

6 0
3 years ago
In which of these compounds are there twice as many oxygen atoms as hydrogen atoms?
Deffense [45]

Answer:

G - H2SO4

Explanation:

two hydrogen atoms and 4 oxygen atoms

8 0
3 years ago
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