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andriy [413]
4 years ago
7

Which type of space probe is designed to explore several planets or moons from space as it travels past them?

Chemistry
1 answer:
Doss [256]4 years ago
8 0
C) the term"orbit' means to revolve around a certain object thus the term to orbit is the most accurate option in witch there is to choose!
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Draw a line-bond structure for CBrN. Explicitly draw all H atoms. You do not have to include lone pairs in your answer. In cases
DochEvi [55]

Answer:

Br - C ≡ N

Explanation:

To draw the Lewis line-bond structure we need to bear in mind the octet rule, which states that in order to gain stability each <em>atom tends to share electrons until it has 8 electrons in its valence shell</em>.

  • C has 4 e⁻ in its valence shell so it will form 4 covalent bonds.
  • Br has 7 e⁻ in its valence shell so it will form 1 covalent bond.
  • N has 5 e⁻ in its valence shell so it will form 3 covalent bonds.

The most stable structure that respects these premises is:

Br - C ≡ N

It does not have any H atom.

8 0
3 years ago
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
Please help, I’m a bit confused, lol
Mashutka [201]

Explanation:

option A skeleton system is correct option

hope this helps you !

4 0
2 years ago
I HAVE ANSWER
Agata [3.3K]
<span>11.2G is the answer to this problem.
</span>
7 0
3 years ago
Read 2 more answers
Which alcohol will undergo acid-catalyzed dehydration under the mildest conditions?
damaskus [11]
<span>Tertiary alcohols are the type of alcohols that will undergo acid-catalyzed dehydration under the mildest conditions. Types of tertiary alcohols are 2-methylpropan-2-ol and 2-methylbutan-2-ol. Other types of alcohols are referred to as primary alcohols and secondary alcohols.</span>
6 0
3 years ago
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