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Artyom0805 [142]
3 years ago
9

HELP!!!! please omg

Chemistry
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer:

I believe it's A. to reduce air bubbles. Tbh, it's been a while

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Suppose 'A' is a liquid aromatic compound with molecular weight 78 and burns with sooty flame. a.Give the name of the compound '
Oliga [24]

Aromatic compounds are compounds that contain carbon-carbon multiple bonds.

The question did not mention that a heteroatom is present in the compound so we can assume that there is none of such. In that case, the compound contains only hydrogen and carbon.

So,

(CH)n = 78

where n is the number of each atom present.

(12 +1)n = 78

n = 78/13

n = 6

The molecular formula of the compound is C6H6

When C6H6 is treated with .conc.HNO3/conc.H2SO4 the compound shown in image 1 is formed. The reaction occurs at the C-C multiple bond.

When C6H6 is reacted with chlorine in the presence of sunlight, hexachlorobenzene (shown in image 2 attached) is formed.

brainly.com/question/24305135

3 0
2 years ago
The chemical reaction that breaks the bonds between chains of monomers is called?
Naily [24]
Hydrolysis !  is the answer to your question!
4 0
3 years ago
The energy gap between the valence band and the conduction band in the widely-used semiconductor gallium arsenide (GaAs) is Δ =
stira [4]

Answer:

n_e = 8.139 *10^6

Explanation:

Given that;

The energy gap between the valence band and the conduction band in the widely-used semiconductor gallium arsenide (GaAs) is Δ = 1.424 eV.

So; that implies that:

\delta \ E = 1.424 \ eV

Suppose that we consider a small piece of GaAs with 1020 available electrons, -- This is taking about the numbers of electrons used which is :

n_i = 10^{20} \ electrons

Temperature is given as:

T = 274.15 \ K

Number of electrons can be calculated by using the formula;

n_e = n_i*e^-^{\frac{\delta E }{2 K_B*T}}

n_e = 10^{20}*e^-^{\frac{1.424}{2*8.617*10^{-5}*274.15}}

n_e = 8.139 *10^6

8 0
2 years ago
I need help with this asap.​
Vanyuwa [196]

Answer:

H30+ First H20 Second N03 Third HNO3 Fourth

Explanation:

4 0
2 years ago
Reynolds number E. What is the mean velocity u. (ft/s) and the Reynolds number Re = pu., D/ for 35 gpm (gallons per minute) of w
Novay_Z [31]

Answer:

The mean velocity is 13 ft/s.

The Reynolds number is 88,583 and it is dimensionless.

Explanation:

We have water flowing in a pipe of 1.05 in diameter.

The density is ρ=62.3 lb/ft and the viscosity is 1.2 cP.

The mean velocity can be calculated as

u=\frac{Q}{A}=\frac{Q}{\pi*D^2/4}=\frac{35gpm }{3.14*(1.05in)^2/4}\\\\  u=\frac{35}{0.865}*\frac{gal}{min}\frac{1}{in^2}*\frac{231in^3}{1gal}*\frac{1}{60s} \\\\    u=156\,in/s=13\,ft/s

The Reynolds number now can be calculated for this flow as

Re=\frac{\rho*u*D}{\mu}

being ρ: density, u: mean velocity of the fluid, D: internal diameter of the pipe and μ the dynamic viscosity.

To simplify the calculation, we can first make all the variables have coherent units.

<em>Viscosity</em>

\mu=1.2cP=\frac{1.2}{100}\frac{g}{cm*s}*\frac{1lb}{453.6g}*\frac{30.48cm}{1ft}= 0.0008\frac{lb}{ft*s}

<em>Diameter</em>

D=1.05in*(\frac{1ft}{12in} )=0.0875ft

Then the Reynolds number is

Re=\frac{\rho*u*D}{\mu}\\\\Re=62.3\frac{lb}{ft^3}*13\frac{ft}{s} *0.0875ft*\frac{1}{0.0008}*\frac{ft*s}{lb}\\\\Re=88,583

8 0
3 years ago
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