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puteri
8 days ago
7

Pyridine rings can also under electrophilic aromatic substitution. given 2-methoxypyridine below, draw the expected major produc

t by adding in the new substitution.
Chemistry
1 answer:
alisha [2.6K]8 days ago
8 0
Co2 is indeed the correct answer, my friend.
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When performing a flame test using the method described in the manual, you complete the flame test of KNO3 and find a yellow col
Tems11 [2321]

Answer:

The nichrome wire has contaminants.

The sample solution might be tainted.

Explanation:

If the nichrome wire is contaminated, sodium impurities could be causing the yellow flame. The wire is initially placed in the flame without the sample to check for such impurities.

The testing solution could also be contaminated, causing it to display a color different from the anticipated shade of the test ion.

7 0
1 month ago
Acetone major species present when dissolved in water
Alekssandra [2611]

The compound is acetone ( CH₃-CO-CH₃)


Explanation:


1) Acetone is represented as CH₃-CO-CH₃.


2) This is a molecule formed by covalent bonds.


3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.


In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.


This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).



3 0
1 month ago
Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi
Tems11 [2321]

Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

7 0
1 month ago
A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,
Anarel [2530]

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

Based on Archimedes' principle, the mass of fresh water and the mass of the cup are equal to the mass of the same volume of seawater.

The mass of freshwater can be calculated using density times volume.

1 cm³ is equivalent to 1 mL.

The mass of freshwater is 0.999 g/cm³ multiplied by 735 cm³, which results in 734.265 g.

The total mass of the freshwater and cup combined is 734.265 g plus 25 g, equating to 759.265 g.

This means the mass for an equal volume of seawater is 759.265 g.

The volume of the seawater displaced is 735 mL, which is 0.735 L (assuming the cup's volume can be disregarded).

We know that 1 liter equals 1000 cm³ or 1000 mL.

The density of seawater can be determined as mass divided by volume.

The density of seawater becomes 759.265 g divided by 0.735 L, yielding 1033.01 g/L.

Conversely, the density of freshwater in g/L is calculated as 0.999 g/(1/1000) L, equating to 999 g/L.

The mass of salt dissolved in 1 liter of seawater is calculated as 1033.01 g - 999 g, which equals 34.01 g.

Thus, the amount of salt in 1 L of seawater is 34 g.

4 0
1 month ago
In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a
KiRa [2606]

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

\Delta H =\frac{Q}{n}

To calculate the heat (Q):

Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J

Next, we need to calculate the number of moles involved in the reaction (n):

n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles

With these two values, we can substitute them into the first equation:

\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole

Have a great day!

5 0
1 month ago
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