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4 years ago

Is c6h12o6 an empirical formula?

Chemistry

1 answer:

ziro4ka [17]4 years ago

5 0

Answer:

the molecular formula of glucose, C6H12O6 = 6 x CH2O c. The molecular weight of glucose is 180 g/mol. It is equal to 6 x 30 g/mol. g/mol, so the empirical and molecular formulas are the same.

Explanation:

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3 enzimas presentes en nuestro organismo e indique en que procesos actúan

lapo4ka [179]

Answer:

ATP asa, Helicasa, Proteasa, ARN polimerasa

Explanation:

Las enzimas son un tipo de biomoleculas que se corresponden con las proteinas.

Al momento de referirse a ellas, se utiliza la terminación asa.

ATPasa → Sintetizando ATP para el funcionamiento celular

Helicasa → Abre las hebras de ADN permitiendo el paso de la horquilla para el proceso de replicación de ADN.

Proteasas → Enzimas que degradan proteinas mal plegadas, rompen los enlaces peptídicos.

ARN polimerasa → Sintesis de ARN mensajero a partir de ADN en el proceso de la Transcripción. Se la puede conocer a veces, como primasa.

8 0

3 years ago

Consider this equilibrium: HCO31- + H2PO41- HPO42- + H2CO3. What are the Brønsted-Lowry acids in this equilibrium?

horrorfan [7]

To me i think the answer is c!

4 0

3 years ago

Read 2 more answers

Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B

Ivenika [448]

Answer:

-2546 kJ

Explanation:

It is possible to obtain the enthalpy of a reaction from the sum of different intermediate reactions.

For the reaction:

4B(s) + 3O₂(g) → 2B₂O₃(s)

The intermediate reactions are:

A- B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g), ΔH°A= +2035 kJ

B- 2B(s) + 3H₂(g) → B₂H₆(g), ΔH°B= +36 kJ

C- H₂(g) + 1/2O₂(g) → H₂O(l), ΔH°C= -285 kJ

D- H₂O(l) → H₂O(g), ΔH°D= +44 kJ

2B = 4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH°2B= +78 kJ

-2A = 6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH°-2A= -4070 kJ

-6C = 6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH°-6C= +1710 kJ

-6D = 6H₂O(g) → 6H₂O(l) ΔH°-6D = -264 kJ

The sum of 2B - 2A - 6C - 6D produce:

4B(s) + 3O₂(g) → 2B₂O₃(s)

And the enthalpy is: ΔH°2B + ΔH°-2A + ΔH°-6C + ΔH°-6D = -2546 kJ

I hope it helps!

3 0

3 years ago

Assume that the partial pressure of sulfur dioxide, pso2, is equal to the partial pressure of dihydrogen sulfide, ph2s, and ther

Alex Ar [27]

Kp = PH2O^2/(PH2O)^2 To calculate SO2 is to have a Kp for the reaction let x=PSO2=PH2S Kp= (28/760atm)^2/x^3 x=cube root(0.03684)/Kp=PSO2

8 0

4 years ago

2-- What is the [H3O+] in a solution with [OH-] = 1 x 10-12 M?

Arte-miy333 [17]

Answer:

The answer is 0.01 M

Explanation:

The problem is solved by applying the expression for ionic product of water as follows:

Kw = [H₃O⁺] [OH⁻]

Where Kw is the ionic product of water and it is 1.10⁻¹⁴ M at⁴ 25ºC. As [OH⁻]= 1.10⁻¹² M, [H₃O⁺] will be:

[H₃O⁺]= Kw/ [OH⁻]= 1.10⁻¹⁴M/1.10⁻¹²M= 0.01 M

5 0

3 years ago