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labwork [276]
4 years ago
9

During _______ materials move into the cell while during ______ materials move out of the cell.

Chemistry
1 answer:
Len [333]4 years ago
3 0
D. Endocytosis ; Exocytosis
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Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of SO2 by O2 to give SO3
SIZIF [17.4K]

Answer:

See explanation

Explanation:

We know that the process of the oxidation of SO2 to SO3 is catalysed by NO2 gas. It occurs in two stages and i will show the balanced reaction equation of the both stages below;

Step 1

2NO2(g) + 2SO2(g) ------> 2NO(g) + 2SO3(g)

Step 2

2NO(g) + O2(g) -------> 2NO2(g)

So, the overall reaction equation is;

2SO2(g) + O2(g) ------> 2SO3(g)

5 0
3 years ago
Predict the aldol product when the following ketone undergoes self-condensation in the presence of NaOH. Do not show the dehydra
Anit [1.1K]

Answer:

β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal.

Explanation:

When acetaldehyde is treated with dil.NaOH it undergoes self condensation as it contains alpha-hydrogen atom in its compound forming β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal. This compound upon further heating will eliminate a molecule of water forming aldol condensation product namely Crotonaldehyde Or But-2-en-al. see the diagram attached.

5 0
3 years ago
What is the name of an acid with the formula HNO2?
Step2247 [10]
HNO2 is the formula for Nitrous Acid
8 0
3 years ago
Control of Blood pH by respiratory rate.
nata0808 [166]

Answer:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in  H^+ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

Explanation:

A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system

CO_{2} +H_{2} O ⇄ H^+ + HCO^-_{3}

B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in  H^+ which is directly proportional to the increase in Blood PH levels

C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid

6 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th
WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0
3 years ago
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