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Elenna [48]
3 years ago
14

A student tests the acidity of nitric acid (HNO3) by dissolving a sample of HNO3 in a solution of liquid methane. He then uses a

n electronic tool to measure the solution and determines that it is highly acidic. What is wrong with the student’s experiment?
Physics
2 answers:
DerKrebs [107]3 years ago
8 0
Sorry I didn't see this before...
Okay, I see two major problems with this student's experiment:

1) Nitric acid Won't Dissolve in Methane
Nitric acid is what's called a mineral acid. That means it is inorganic (it doesn't contain carbon) and dissolves in water.
Methane is an organic molecule (it contains carbon). It literally cannot dissolve nitric acid. Here's why:
For nitric acid (HNO3) to dissolve into a solvent, that solvent must be polar. It must have a charge to pull the positively charged Hydrogen off of the Oxygen. Methane has no charge, since its carbon and hydrogens have nearly perfect covalent bonds. Thus it cannot dissolve nitric acid. There will be no solution. That leads to the next problem:

2) He's Not actually Measuring a Solution
He's picking up the pH of the pure nitric acid. Since it didn't dissolve, what's left isn't a solution—it's like mixing oil and water. He has groups of methane and groups of nitric acid. Since methane is perfectly neutral (neither acid nor base), the electronic instrument is only picking up the extremely acidic nitric acid. There's no point to what he's doing.

Does that help?
tester [92]3 years ago
8 0

Answer:

B

Explanation:

He dissolved his sample in methane instead of water.

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3 years ago
The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit
Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V=  30 m/s  ( + x direction )

d)  λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

As we know that standard form of wave equation given as

y=A sin(\phi -\omega t)

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s                     (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s  ( + x direction )

V = f λ

30 = 95.49 x  λ

 λ = 0.31 m

We know that speed is the rate of displacement

U=\dfrac{dy}{dt}

U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0
3 years ago
A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is us
Nataliya [291]

Answer:

T_C=118.8 K= 154.2°C

Explanation:

COP_max of carnot heat pump= \frac{T_{H} }{T_{H}-T_{C} }

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP=\frac{Q_H}{W}

in the question COP= \frac{60}{100} \times COP_{max}

⇒\frac{Q_H}{W} =\frac{60}{100}\times\frac{T_H}{T_H-T_C}

heat is added directly to be as efficient as via heat pump

Q_H= W

and T_H= 24° C= 297 K

1=\frac{60}{100}\times \frac{297}{297-T_C}

on calculating the above equation we get

T_C=118.8 K

the outdoor temperature for efficient addition of heat to interior of home

T_C=118.8 K= 154.2°C

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Complete the statement by filling in the Blanks:
Anika [276]

Answer:

induced current

Explanation:

intentionally manipulated.

5 0
3 years ago
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