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3 years ago

All molecular motion stops at 0 on this scale

Chemistry

2 answers:

Oxana [17]3 years ago

4 0

Kelvin Hope this helps

maxonik [38]3 years ago

4 0

The answer is Kelvin

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What is the measurement for momentum?

sertanlavr [38]

The unit of momentum is the product of the units of mass and velocity.

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3 years ago

A mixture is made of 40 ml of salt water to 200 ml of solution. What percent of the solution is salt water?

Serggg [28]

Answer:

16.7%

Explanation:

40 ml of salt water + 200 ml of solution = 240 ml

40/240 = 4/24 = 1/6=16.7%

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3 years ago

A pure gold bar is made up of only particles. A.nickel B.copper C.god D.gold

Minchanka [31]

Answer: gold

Explanation:

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3 years ago

A more electronegative atom A) will have more attraction to the electrons in a chemical bond. B) is more likely to lose an elect

DIA [1.3K]

Answer: A more electronegative atom will have more attraction to the electrons in a chemical bond.

Explanation:

An atom that is able to attract electrons or shared pair of electrons more towards itself is called an electronegative atom.

For example, fluorine is the most electronegative atom.

Due to its high electronegativity it is able to attract an electropositive atom like H towards itself. As a result, both fluorine and hydrogen will acquire stability by sharing of electrons.

Thus, we can conclude that a more electronegative atom will have more attraction to the electrons in a chemical bond.

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

rodikova [14]

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

2 years ago