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3 years ago

Which element was first discovered in space before it was discovered on Earth?

2 answers:

8 0

Helllllooo Friend __________________________:-) :-):-) :-)

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Helium is named from the Greek helios, "the Sun", where the element was discovered through spectroscopy.

Hope this answer help you!!!!!!!!!!! :-) :-) :-) :-) :-) :-)

faltersainse [42]3 years ago

7 0

Helium was first discovered in space before it was on earth

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The lock-and-key mechanism refers to

miskamm [114]

Answer:

A). The complementary shapes of an enzyme and a substrate.

Explanation:

The Lock-and-key mechanism was proposed by Emil Fischer for the first time and characterized as the metaphor which helps in elucidating the specificity of the enzymatic reactions. In this metaphor, the lock is described as the enzyme while 'key' is characterized as the substrate which the enzyme acts upon. If the key is not appropriately sized, it will not fit into the active site i.e. the keyhole of the lock or enzyme and reaction will not take place. Thus, <u>option A</u> is the correct answer.

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3 years ago

Read 2 more answers

Pure water is neutral. if water is neither an acid nor base what is the ph of the water

SCORPION-xisa [38]

Its pH is constant which is neither acid or base

7 0

4 years ago

Read 2 more answers

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

A 133.8-gram sample of bronze was 10.3% tin by mass. Determine the total mass of tin in the sample.

Kay [80]

You just need to multiply the total mass by the decimal value of the part that is tin. 133.8*0.103=13.8g (following the rules of significant figures).

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4 years ago

How many valence electrons are in a stable ion of a period 4 metal?

marusya05 [52]

What kind of metal is it

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3 years ago