The hormone adrenaline can affect only cells with

3241004551 [841]

Answer:I can't see any question

8 0

3 years ago

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 CO2 produced in one round of the citric acid cycle does not originate in the acetyl carbons that entered that round. If the

aleksley [76]

Answer:

It will require<u> second round</u> of the cycle to release $14C0_2$

Explanation:

<u>Reason behind the requirement of second round of the cycle to release </u> $CO_2$ -:

The C4 carbon of succinyl CoA is acetyl from acetyl CoA. Succinyl CoA is converted to succinate, which is then converted to fumarate, fumarate, malate, and eventually oxaloacetate. 14C will be found in oxaloacetate at either C1 or C4. During the second round of the loop, each of these carbons will be converted to carbon dioxide.

7 0

2 years ago

6.0 mol NaOH reacts with

lina2011 [118]

Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

NaOH: 3 moles
H₃PO₄: 1 mole
H₂O: 3 moles
Na₃PO₄: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

$moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}$

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

$moles of Na_{3}P O_{4} =\frac{6 moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}$