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White raven [17]
3 years ago
11

Mass and energy are alternate aspects of a single entity called mass-energy. The relationship between these two physical quantit

ies is Einstein's equation, E = mc2, where E is energy, m is mass, and c is the speed of light. In a combustion experiment, it was found that 10.965 g of hydrogen molecules combined with 87.000 g of oxygen molecules to form water and released 1.554 × 103 kJ of heat. Use Einstein's equation to calculate the corresponding mass change in this process.
Physics
1 answer:
Sever21 [200]3 years ago
8 0

Answer:

Explanation:

ΔE = Δm × c^2

where,

ΔE = change in energy released with respect to change in mass

= 1.554 × 10^3 kJ

= 1.554 × 10^6 J

Δm = change in mass

c = the speed of light.

= 3 × 10^8 m/s

Equation of the reaction:

2H2 + O2 --> 2H2O

Mass change in this process, Δm = 1.554 × 10^6/(3 × 10^8)^2

= 1.727 × 10^-11 kg

The change in mass calculated from Einstein equation is small that its effect on formation of product will be negligible. Hence, law of conservation of mass holds correct for chemical reactions.

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valentina_108 [34]

Answer:

It is 10.75

Explanation:

8 0
3 years ago
Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
5 0
4 years ago
What is the de Broglie - equation​
fomenos

Answer:

is this it?

Explanation:

λ = h/mv, where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.

8 0
3 years ago
A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of
iogann1982 [59]

Answer:

8.64283\times 10^{-14}\ N

Explanation:

q = Charge of proton = 1.6\times 10^{-19}\ C

v = Velocity of proton = 0.267\times c

c = Speed of light = 3\times 10^8\ m/s

B = Magnetic field = 0.00687 T

\theta = Angle = 101^{\circ}

Magnetic force is given by

F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N

The magnetic force acting on the proton is 8.64283\times 10^{-14}\ N

4 0
4 years ago
If the velocity of the incident charged particle were doubled, how would b have to change (keeping all other parameters constant
Paraphin [41]
If you double the velocity you would have to halve the magnetic field strength to keep the force on the particle the same and hence the trajectory.

F = qvB
6 0
3 years ago
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