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8090 [49]
3 years ago
11

An oxidizing flame resembles the neutral flame slightly, but has an inner cone that is shorter and more pointed with an almost p

urple color instead of brilliant white true or false
Chemistry
1 answer:
jok3333 [9.3K]3 years ago
3 0

Answer:

True

Explanation:

The answer to this question is true. The oxidizing flame is a flame which is gotten when oxygen is used excessively. The flame is bluish or almost purple in color. And it makes a hissing sound. The flame from this is shorter than the neutral flame. In the neutral flame there is neither the occurrence of oxidation or reduction.

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Write the molecular formula of sodium bicarbonate and write any two uses of it.
gulaghasi [49]

\huge\mathcal\colorbox{cyan}{{\color{blue}{AnSwEr~↓~↓~}}}

\huge\mathbb\colorbox{black}{\color{white}{FORMULA\:\:↓}}

<h3><em>NaHCO3</em></h3>

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5 0
3 years ago
2-Ethoxy-2,3-dimethylbutane reacts with concentrated aqueous HI to form two initial organic products (A and B). Further reaction
Leno4ka [110]

Answer:

A = 2-iodo-2,3-dimethylbutane

B = Ethanol

C = Iodoethane (also called ethyl-iodide)

Explanation:

2-Ethoxy-2,3-dimethylbutane reacts with conc. HI to cleave the oxy-functional group.

On one end, ethanol is formed and on the other hand, 2-iodo-2,3-dimethylbutane is formed.

But ethanol reacts further with conc HI to give iodoethane.

Therefore,

A = 2-iodo-2,3-dimethylbutane

B = Ethanol

C = Iodoethane (also called ethyl-iodide)

This is all shown in the attached image.

Hope this Helps!!!

6 0
4 years ago
Thermodynamic PropertiesProperty Value
igomit [66]

Answer:

1,620 J.

Explanation:

  • The amount of heat added to a substance (Q) can be calculated from the relation:

<em>Q = m.c.ΔT.</em>

where, Q is the amount of heat released from ethanol cooling,

m is the mass of ethanol (m = 60.0 g),

c is the specific heat of ethanol in the liquid phase, since the T is cooled below the boiling point and above the melting point (c = 1.0 J/g °C),

ΔT is the temperature difference (final T - initial T) (ΔT = 43.0 °C – 70.0 °C = - 27.0 °C).

<em>∴ Q = m.c.ΔT</em> = (60.0 g)(1.0 J/g °C)(- 27.0 °C) = - 1620 J.

<em>The system releases 1620 J.</em>

8 0
3 years ago
Find the number of molecules and number of each type of atom of 0. 04 moles of H2C2O4
Galina-37 [17]

answer

Explanation: i think 0.49

3 0
2 years ago
Calculate the percent dissociation of trimethylacetic acid(C6H5CO2H) in a aqueous solution of the stuff.
Sophie [7]

Answer:

1.112%.

Explanation:

Step one : write out the correct acid dissociation reaction. This is done below:

C6H5CO2H <=====> C6H5CO2^- + H^+.

At equilibrium, the concentration of C6H5CO2H = 0.5 M - x and the concentration of C6H5CO2^- = x and the concentration of H^+ = x.

The ka for C6H5CO2H = 6.3 × 10^-5.

Step two: find the value for the concentration of C6H5CO2^- and H^+. This can be done by using the equilibrium dissociation Constant formula below;

Ka = [C6H5CO2^- ] [H^+] / C6H5CO2H.

ka = [x] [x] / [0.5 - x].

6.3 × 10^-5 = (x)^2 / [0.5 - x].

x^2 = 3.15 ×10^-5 - 6.3 × 10^-5 x.

x^-2 + 6.3 × 10^-5 x - 3.15 × 10^-5.

Solving for x we have x= 0.0055632289702004 = 0.00556.

Therefore, at equilibrium the concentration of H^+ = 0.00556 M and the concentration of C6H5CO2H= O.5 - 0.00556 = 0.494 M.

Step three: Calculate the equilibrium pH. This stage can be ignored for this question since we are not asked to calculate for the pH.

The formular for pH = - log [H^+].

pH= - log [0.00556].

pH= 2.255.

Step four: find the percentage dissociation.

Percentage dissociation = ( [H^+] at equilibrium/ [ C6H5CO2H] at Initial concentration ) × 100%.

Percentage dissociation=( 0.00556/ 0.5 ) × 100.

=Percentage dissociation= 1.112%.

3 0
3 years ago
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