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KiRa [710]
3 years ago
12

The MSDS for glacial acetic acid says that it is a flammable liquid that can severely burn any human tissue it comes in contact

with. It reacts with bases, various metals, and strong oxidizing agents. Its vapors can form explosive mixtures with air. Based on this information, which statement is applicable to glacial acetic acid?
Physics
1 answer:
Alex777 [14]3 years ago
3 0

Answer: D) It should be handled in a fume hood, away from open flames.

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Un chico de 45 kg de masa se encuentra de pie sobre la nieve. Calcula la presión sobre este:
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6 0
3 years ago
In nuclear fission, a nucleus splits roughly in half. (a) What is the potential 2.00 × 10−14 m from a fragment that has 46 proto
Valentin [98]

Answer:

electric potential is 3.31 × 10^{6} V

potential energy is 152 MeV

Explanation:

given data

fragment charge  Q = 46 protons = 46 × 1.6 × 10^{-19} C

to find out

electric potential  and potential energy

solution

we know here distance from fragment d = 2 × 10^{-14} m

and constant for electric force k that is 9 × 10^{9} N-m²/C²

so that we can find electric potential = kQ/d

electric potential = 9 × 10^{9}[/tex ×46 × 1.6 × [tex]10^{-19} / ( 2 × 10^{-14} )

electric potential = 3.31 × 10^{6} V

and

we know relation between electric potential and potential

that is  V = U/q

so U will be = qV

now put all value

we get potential energy U

potential energy = 46 × 3.31 × 10^{6}

potential energy = 1.52 × 10^{8} eV

so potential energy = 152 MeV

4 0
3 years ago
A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the
fredd [130]

Answer:

  t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material  = 1.375

we know the equation

       c = ν λ

where ν is the frequency of the wave

           c is the speed of light

   \nu= \dfrac{c}{\nu\lambda}

   \nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}

       ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

        t = \dfrac{\lambda}{4\mu_{material}}

        t = \dfrac{522 \times 10^{-9}}{4\times 1.375}

              t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm

7 0
3 years ago
30 POINTS!
brilliants [131]

Answer:

d

Explanation:

5 0
3 years ago
Read 2 more answers
considere que o calor específico de um material presente nas cinzas seja c=0,8j/gc. Supondo que esse material entre na turbina a
drek231 [11]

Answer:

3120J

Explanation:

Given parameters:

C  = Specific heat capacity  = 0.8J/g°C

Initial temperature  = 20°C

Mass given   = 5g

Final temperature  = 800°C

Unknown:

Energy given to the mass  = ?

Solution:

To find the energy given to the mass, let us simply use the expression below:

          H   =   m   c   ΔT

H is the unknown, the energy supplied

m is the mass of the substance

c is the specific heat capacity

ΔT is the change in temperature

Input the variables;

            H    = 5  x   0.8    x    (800 - 20)  = 3120J

7 0
3 years ago
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