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3 years ago

Can a electron be found in an exact spot within a atom

Chemistry

1 answer:

goldenfox [79]3 years ago

5 0

Answer:

Electrons are located in an electron cloud, which is the area surrounding the nucleus of the atom. There is usually a higher probability of finding an electron closer to to the nucleus of an atom.

Explanation:

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What is one of the primary unanswered questions about the origin of the universe? What type of elements formed first? Approximat

torisob [31]

Answer: last option, what came before the big bang?

Explanation:

The big bang theory states that the universe started as a dense nucleus of matter: a huge amount of matter concentrated in a tiny spot.

This is the conclusion of equations and evidences that prove that the universe has been and continuous to expand: since it has been expanding, there was a moment when it was as small and dense as it is possible.

So, the expansion is the result of violent explosion.

The time during which the expansion has been happening (this is how long ago the big bang occured) has been estimated thanks the the observation of the speed of recesion of the galaxies, but nothing can be told about what came before the bing bang occured.

7 0

3 years ago

Read 2 more answers

A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

3 0

3 years ago

Cu + 2AgNO3 →

netineya [11]

Answer: 292.54g of Ag

Explanation:

Cu + 2AgNO3 →Cu(NO3)2 + 2 Ag

mass conc. Of Ag = n x molar Mass

Mass conc. Of Ag = 2 x 108 = 216g

From the equation,

63.5g of Cu produced 216g of Ag

Therefore, 86g of Cu will produce Xg of Ag. i.e

Xg of Ag = (86 x 216)/63.5 = 292.54g

5 0

3 years ago

Which expressions are equal to 675? 700 – 25 O 700 – 75 O 600 + 75 O 634 + 31 500 275

ser-zykov [4K]

Answer:

Explanation:

6 0

2 years ago

Read 2 more answers

Carry out the following operations as if they were calculations of experimental results and express each answer in standard nota

Shalnov [3]

Answer:

1. $10.6\; \rm m$ (one decimal place.)

2. $0.79\; \rm g$ (two decimal places.)

3. $16.5\;\rm cm^2$ (three significant figures.)

Explanation:

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

$5.6792\;\rm m$ has four decimal places.
$0.6\; \rm m$ has only one decimal place.
$4.33\; \rm m$ has two decimal places.

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: $\rm m$ .)

Therefore:

$\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m$ . (Rounded to one decimal place.)

Similarly:

$\rm 3.70\; \rm g$ has two decimal places.
$2.9133\; \rm g$ has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be $\rm g$ (same as the unit of the two inputs.)

$\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g$ . (Rounded to two decimal places.)

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

$4.51\; \rm cm$ has three significant figures.
$3.6666\; \rm cm$ has five significant figures.

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be $\rm cm \cdot cm$ , which is occasionally written as $\rm cm^2$ .

$\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2$ . (Rounded to three significant figures.)

7 0

3 years ago