How does an atom change if all of its electrons are removed

Physics

2 answers:

Makovka662 [10]3 years ago

4 0

It becomes an ion, depending on how many protons the atom has

worty [1.4K]3 years ago

4 0

It becomes positively charged since the electrons are removed and it will be left with protons which are positively charged.

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Assigned Media Question Help According to a random sample taken at 12 A.M., body temperatures of healthy adults have a bell-sh

deff fn [24]

Answer:

$1- \frac{1}{2^2} = 1-\frac{1}{4} = \frac{3}{4}$

So then we will have at least 75% of the data within two deviations from the mean $\bar x \pm 2s$ .

$\bar x -2s = 98.32 -2*0.59 = 97.14$

$\bar x +2s = 98.32 +2*0.59 = 99.5$

So then the interval would be (97.14, 99.5)

Explanation:

Previous concepts

Chebyshev’s rule is appropriate for any distribution. "Chebyshev’s inequality applies to all distributions, regardless of shape". And is useful since provides a "minimum percentage of the observations that lies within k standard deviations of the mean. "

If k = 2, at least 3/4 of the measurements lie within 2 standard deviations to within the mean.

And the general formula is (1-1/k^2) represent the fraction of the data within the mean .

Solution to the problem

For this case we want to find the percentage of data that would be at least within two deviations from the mean so for this case the value of k =2 and if we replace we got:

$1- \frac{1}{2^2} = 1-\frac{1}{4} = \frac{3}{4}$

So then we will have at least 75% of the data within two deviations from the mean $\bar x \pm 2s$ .

For the other part we have the mean and deviation provided $\bar x = 98.32$ $s = 0.59$ the interval would be:

$\bar x -2s = 98.32 -2*0.59 = 97.14$

$\bar x +2s = 98.32 +2*0.59 = 99.5$

So then the interval would be (97.14, 99.5)

6 0

4 years ago

Please need help on this

Vera_Pavlovna [14]

Answer:

i have no idea

Explanation:

it could be B. but i dont belive in myself sooooooooooo.....ehhhh yeah

4 0

3 years ago

The start-up procedure for a batch reactor includes a heating step where the reactor temperature is gradually heated to the nomi

qwelly [4]

Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)