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KonstantinChe [14]
3 years ago
14

What is the melting point and boiling point of Gold in Celsius and what is the # of protons and Neutrons of it oh and the number

of electrons?
Chemistry
1 answer:
nikitadnepr [17]3 years ago
7 0
The  of melting point of gold is 1,948 degrees F and 1,064 degrees C 
<span>The boiling point is 4,892 degrees F and 2,700 degrees C

Yet the numbers neutrons are 12. The number of protons is 36 and electrons is the same as protons 

hope this helped :)

</span>
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glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate AT
ddd [48]

Answer:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula:  ∆Go ’= -RTln K’eq

where,  

R = -8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 270

=  - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 890

=  - 8.315 x 298 x 6.79

=  - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

=  -30.7 kj/mol

Thus, -30.7 kj/mol is the correct answer.

6 0
3 years ago
Examples of two substances being mixed and the ​
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Answer:

ocean water liquid and steel solid

6 0
3 years ago
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
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Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
3 years ago
Write the electron configuration for the element titanium, Ti.
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Answer:

[Ar] 3d^{2} 4s^{2}

Explanation:

You can look at the periodic table and figure out the electron config.

8 0
3 years ago
A sample of no2 occupies 16,500 ml at stp. How many no2 molecules are in the sample?.
svet-max [94.6K]
here it is! mL to Liters to moles to molecules

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2 years ago
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