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balandron [24]
3 years ago
13

In the following reaction, which compounds are considered based? HNO2+H2O <—>H3O + NO2

Chemistry
1 answer:
ehidna [41]3 years ago
7 0

Answer:

non of them are considered base hno3 is acidic ,H2O is neutral ,h30 is acidic likewise n02

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The lock-and-key mechanism refers to
miskamm [114]

Answer:

A). The complementary shapes of an enzyme and a substrate.

Explanation:

The Lock-and-key mechanism was proposed by Emil Fischer for the first time and characterized as the metaphor which helps in elucidating the specificity of the enzymatic reactions. In this metaphor, the lock is described as the enzyme while 'key' is characterized as the substrate which the enzyme acts upon. If the key is not appropriately sized, it will not fit into the active site i.e. the keyhole of the lock or enzyme and reaction will not take place. Thus, <u>option A</u> is the correct answer.

5 0
3 years ago
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The burning of gasoline to power internal combustion engines produces water vapor, and carbon dioxide gas that is thought to con
ser-zykov [4K]

Answer:

the balanced equation to the combustion of gasoline is C8H18 + 12.5 O2 → 8 CO2 + 9 H2O (1)

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3 years ago
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8.7 Two products are formed in the following reaction in a 50:50 mixture. Would the resulting solution be optically active
k0ka [10]

Answer:

Yes. The solution would be optically active.

Explanation:

Diastereomer are defined as the image that is non mirror and non -identical. It is made up of two stereoisomers. They are formed when the two stereoisomers or more than two stereoisomers of the compound have the same configuration at the equivalent stereocenters.

In the given context, as the product given is a diastereomeric mixture, the product would have an optical activity in total.

So the answer is Yes.

4 0
2 years ago
When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu
IgorLugansk [536]

When connectors are marked with a combination of metals, it can be used as a connector of one of the metals or an alloy of the two metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and Cu is copper, therefore the answer is:

<span>Yes, it is suitable for use with copper, copper-clad aluminum, and aluminum conductors.</span>

6 0
3 years ago
How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?
meriva

Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

                Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

    V/mL:     249

c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}

3. Calculate the volume of HCl

V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}

8 0
3 years ago
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