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d1i1m1o1n [39]
3 years ago
10

True or false

Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
8 0

Answer:

True.

But it only changes in physical change.

How?

Explanation:

The chemical reaction produces a new substance with new and different physical and chemical properties. Matter is never destroyed or created in chemical reactions. The particles of one substance are rearranged to form a new substance.

In a physical change, a substance's physical properties may change.

A chemical change is a permanent change. A Physical change affects only physical properties i.e. shape, size, etc. ... Some examples of physical change are freezing of water, melting of wax, boiling of water, etc. A few examples of chemical change are digestion of food, burning of coal, rusting, etc.

Hope this helps!

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Calculate the maximum KE and velocity of an electron from zinc by a 275nm photon
Dmitry [639]

Answer:

Kinetic energy: 6.024*10^{-20}

Velocity: 3.64*10^{5}

Explanation:

From the equations of the photo-electric effect,

We know:

hv=hv_0+K.E

Where,

1.h is the Planck's constant which is 6.626*10^{-34}

2.v,v_0 are the frequency of light emitted and threshold frequencies respectively.

3.K.E is the kinetic energy of the electrons emitted.

By fact, we come to know that the threshold frequency of Zn is 300nm

And also v=\frac{c}{d}

Where ,

1. c is the speed of light =3*10^8

2.d is the wavelength.

Thus,

\frac{hc}{d}=\frac{hc}{d_0}+K.E\\K.E=\frac{hc}{d}-\frac{hc}{d_0}\\K.E=hc*(\frac{1}{d}-\frac{1}{d_0})\\K.E=6.626*10^{-34}*3*10^8*(\frac{1}{275*10^{-9}}-\frac{1}{300*10^{-9}})\\K.E=6.024*10^{-20}kgm^2s^{-2}

Now to find velocity:

K.E=\frac{1}{2}mv^2\\v^2=1.324*10^{11}\\v=3.64*10^5

8 0
3 years ago
Help me with number 19 and 20 please
Doss [256]
19) it's not balanced because when adding h2 and o2 u get h2o2 not h2o
20) I'm not sure
7 0
3 years ago
Magnesium (Mg) has nine electrons. Which of the following shows the correct electron configuration for an atom of Mg?
lubasha [3.4K]

Answer:last one

Explanation:

6 0
3 years ago
Consider the reaction 5Br−(aq)+BrO−3(aq)+6H+(aq)→3Br2(aq)+3H2O(l) The average rate of consumption of Br− is 1.20×10−4 M/s over t
dexar [7]

Answer:

0.720 M/s

Explanation:

Let's consider the following balanced equation.

5Br⁻(aq) + BrO⁻₃(aq) + 6H⁺(aq) → 3 Br₂(aq) + 3H₂O(l)

The molar ratio of Br⁻ to Br₂ is 5:3, that is, when 5 moles of Br⁻ are consumed, 3 moles of Br₂ are produced. If the average rate of consumption of Br⁻ is 1.20 × 10⁻⁴ M/s, the average rate of formation of Br₂ is:

\frac{1.20molBr^{-}}{L.s} .\frac{3molBr_{2}}{5molBr^{-}} =\frac{0.720molBr_{2}}{L.s}

3 0
3 years ago
What type of energy conversion takes place in a voltaic cell?
iren2701 [21]
A. is the answer i hope this helps
4 0
3 years ago
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