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2 years ago

describe the trends in the atomic size, location energy and electrongavity from left to right across a period in periodic table

1 answer:

3 0

Atomic size decreases in a period but the ionization energy and electronegativity increases across a period.

<h3>Describe the trends in the atomic size, ionization energy and electronegativity?</h3>

Atomic radius decreases across a period because of nuclear charge increases whereas atomic radius of atoms generally increases from top to bottom within a group because there is again an increase in the positive nuclear charge.

Ionization energy increases when we move from left to right across an period and decreases from top to bottom.

Electronegativity also increases from left to right across a period and decreases from top to bottom.

So we can conclude that atomic size decreases in a period but the ionization energy and electronegativity increases across a period.

Learn more about Electronegativity here: brainly.com/question/24977425

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Define the term hydrate

Salsk061 [2.6K]

Answer:

a compound,typically a crystalline one,in which water molecules are chemically bound to another compound or an element

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2 years ago

Which metal will more easily lose an electron sodium or potassium?

sergey [27]

<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>

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3 years ago

All elements can be classified as metals, metalloids, or ?

svlad2 [7]

All elements can be classified as metals, metalloids, or nonmetals

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3 years ago

A 1.00 liter solution contains 0.31 M hydrocyanic acid and 0.40 M sodium cyanide. If 0.100 moles of barium hydroxide are added t

andriy [413]

Explanation:

acid --> HCN

base --> KCN

now

another base NaOH is added

we know that

base will react with an acid

OH- + HCN ---> CN- + H20

we can see that

HCN is used up , so number of moles of HCN will decrease

CN- is being formed , so number of moles of CN- will increase.

A false

B false

C false

D true

E false

6 0

3 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1

H = $\frac{0.1298}{0.0652}$ = 2

O = $\frac{0.130}{0.0652}$ = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is $CH_2O_2$ .

3 0

3 years ago