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Aleks [24]
1 year ago
9

How does a conjugate acid differ from its conjugate base?

Chemistry
1 answer:
inna [77]1 year ago
3 0

The conjugate acid is differ from its conjugate base as, conjugate acid is formed by strong base whereas conjugate base is formed by strong acid.

conjugate base is differ from conjugate acids by the presence of the proton. The conjugate acid is formed when proton is added to the bases whereas conjugate bases is formed when proton is released by the acids.

Example of corrugate acids are given below.

NH_{3} → NH_{2} ^{-} +H^{+}

In the above example  NH_{2} ^{-} is conjugate acids.

Learn about conjugate acid

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The solubility of glucose at 30°C is
weqwewe [10]

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

400 mL \times \frac{0.996g}{1mL} =398g

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

100gH_2O \times \frac{550gGlucose}{398gH_2O} = 138 gGlucose

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

6 0
3 years ago
How are the three lines of defense the same? They all involve producing fever. They all involve fighting pathogens. They all inv
raketka [301]

Answer:

THEY ALL INVOLVE FIGHTING PATHOGENS

Explanation:

The immune system which is involved in defending the body against infections are diseases involves three lines of defense which are all involved in fighting against pathogens. Pathogens are invaders which when introduced into the body causes harm and therefore makes us sick. The body's first line of defense includes the physical barriers such as the skin, mucous membrane; chemical barriers such as tears, saliva, gastric acid in the stomach. These helps to keep the pathogens from entering the delicate parts of the body and once the pathogens find their way out of the reach of the first line of defense, the second line of defense is initiated. This includes inflammatory effects, swelling, redness, phagocytosis by neutrophils and macrophages. The third line of defense is the actions of lymphocytes which acts on invading microbes. The lymphocytes are of two types; the B and T cells. B cells produces antibodies which fight the antigens and T cells attack the infected cells of the body. There is also the memory cells which keeps information about the invading microbes for future attacks. This enables the body to respond swiftly when next the same type of pathogens attack.

6 0
3 years ago
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goldfiish [28.3K]
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Hope it helped
4 0
3 years ago
Read 2 more answers
Balancing chemical equations
mojhsa [17]

<span>2H2 + O2 → 2H2O</span>

<span>
</span>

<span>okay???</span>

<span>
</span>

4 0
3 years ago
Consider the reaction at 500 ° C 500°C . N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) K c = 0.061 N2(g)+3H2(g)↽−−⇀2NH3(g)Kc=0.06
Sav [38]

Answer:

Q = 0.061 = Kc

Explanation:

Step 1: Data given

Temperature = 500 °C

Kc=0.061

1.14 mol/L  N2

5.52 mol/L H2

3.42 mol/L NH3

Step 2: Calculate Q

Q=[products]/[reactants]=[NH3]²/ [N2][H2]³

If Qc=Kc then the reaction is at equilibrium.  

If Qc<Kc then the reaction will shift right to reach equilibrium.

If Qc>Kc then the reaction will shift left to reach equilibrium.  

Q = (3.42)² / (1.14 * 5.52³)

Q = 11.6964/191.744

Q = 0.061

Q = Kc the reaction is at equilibrium.  

4 0
3 years ago
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