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sesenic [268]
3 years ago
12

A compound consisting of only phosphorous and oxygen atoms is 43.64 % phosphorus by mass.The molecular mass/formula weight of th

e compound is 283.88 amu. How many phosphorous atoms ar in a molecule of this compound? Please show all calc involved.
Physics
1 answer:
DanielleElmas [232]3 years ago
6 0
First, we calculate the mass of Phosphorous present:
283.88 x 0.4364
= 123.88 amu
Atomic mass of P is 31 amu

moles of P = mass / Ar
= 123.88 / 31
= 4.0 moles

We know that one mole of substance has 6.02 x 10²³ particles
Atoms of P = 4 x 6.02 x 10²³ 
= 2.41 x 10²⁴ atoms
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3 years ago
Pressure of gas is 13.6 cm of hg what does it means​
vlada-n [284]

Explanation:

13 cmHg (centimeters of mercury) is the pressure at the bottom of a column of mercury 13 cm deep.  It is the equivalent of about 17.3 kPa or 2.5 psi.

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3 years ago
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

   Hence

        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

7 0
4 years ago
Why construction is present near the bulb ​
vova2212 [387]

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7 0
4 years ago
3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final
MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0
3 years ago
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