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daser333 [38]
3 years ago
10

Assuming that there is a constant partial pressure for oxygen, according to Le Châtelier’s Principle which of the following occu

rs when body temperature decreases during hypothermia?

Chemistry
1 answer:
ololo11 [35]3 years ago
8 0

Answer:

B. The amount of oxygen will decreases

Explanation:

According to Le Châtelier’s Principle, when the condition of the system changes the equilibrium will shift to compensate for the changes. When temperature decrease, the pressure will also decrease and this will shift the equilibrium to the side with exothermic reaction or less molecule count.

The question is not giving the equilibrium reaction of oxygen, so I assume its

O2 + Hb = HbO2

Since the number of molecules in the right side is lower, then the number of oxygen will decrease since the reaction will shift to the right.

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Identify the Lewis acid and Lewis base from among the reactants in each of the following equations.
Alex

Answer:

Part A

Ag+ is the Lewis acid and NH3 is the Lewis base.

Part B

AlBr3 is the Lewis acid and NH3 is the Lewis base.

Part C

AlCl3 is the Lewis acid and Cl− is the Lewis base.

Explanation:

A Lewis acid is any specie that accepts a lone pair of electrons. Ag^+, AlBr3 and AlCl3 all accepted lone pairs of electrons according to the three chemical reaction equations shown. Hence, they are Lewis acids.

A Lewis base donates a lone pair of electrons. They include neutral molecules having lone pair of electrons such as NH3 or negative ions such as Cl- .

3 0
3 years ago
Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha
andrew11 [14]

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

5 0
3 years ago
Use molecular structure and intermolecular bonding to describe why bromine has a lower boiling point than water
kotegsom [21]

Answer:

Bromine mollecules are held together by van der waals forces while a water molecule constitutes both van der waals forces and hydrogen bomnding

Explanation:

This makes the water molecule recquire more heat energy to break the bond thus a higher boiling point while bromine structure requires just litttle heat energy  

3 0
3 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
calculatorn the past, many students have listed that the accidental addition of too much acetic acid contributed greatly to the
Tresset [83]
Is that supposed to be a paragraph
6 0
3 years ago
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