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3 years ago

Where do the electrons come from that produce heat and light in a light bulb?

Chemistry

1 answer:

Sergio [31]3 years ago

8 0

Electrical energy is converted to waste heat. I’m not sure but this is what I think...

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What change in mass number occurs when a radioactive atom emits an alpha particle?

Alik [6]

The alpha particle is the helium nucleus.
The mass of helium nucleus (⁴₂He) is 4u.

The mass of radioactive atom decreases by 4u. Look at the reaction below:

²²³₈₈Ra ---> ⁴₂He + ²¹⁹₈₆Rn

6 0

3 years ago

8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a

sashaice [31]

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

Volume percentage = $\frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100$

= $\frac{88.5}{760} \times 100$

= 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

6 0

3 years ago

Which commercial technology commonly uses plasmas?

Ugo [173]

A television uses plasmas

8 0

3 years ago

Read 2 more answers

I WILL MARK BRAINLIEST ANSWER!

Zina [86]

Answer:

Explanation:

1. They are made up of two or more Pure substances that are not chemically bonded together and appear non-uniform

A heterogeneous mixture is not chemically combined and its components are visible and can be seen.

5 0

3 years ago

Read 2 more answers

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,

Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

$\frac {r^+}{r^-}=0.731$ .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

$r^++r^-=\frac {a}{2}$ ...................2

Given that:

$Cl^-\ (r^-) = 1.82\ \dot{A}$

To find,

$K^+\ (r^+) = ? \dot{A}$

Using 1 and 2 , we get:

$1.731\ r^+=0.731\times \frac {6.28}{2}$

<u>Size of the potassium ion = 1.33 Å</u>

4 0

3 years ago