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3 years ago

9

Why do you think elements have these repeating patterns and groups of similar elements ? Explain your reasoning as much as you c

an.

1 answer:

Evgen [1.6K]3 years ago

6 0

Answer:

The elements in the periodic table are arranged in order of increasing atomic number. All of these elements display several other trends and we can use the periodic law and table formation to predict their chemical, physical, and atomic properties. Understanding these trends is done by analyzing the elements electron configuration; all elements prefer an octet formation and will gain or lose electrons to form that stable configuration.

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Determine the hydrogen ion concentration of solution B. [1]

bazaltina [42]

The hydrogen Ion concentration of solution B is
1.0 x 10^-5 or 0.000 010 M

You can see that this will be proportional to the amount of B's PH compared to A's

hope this helps

8 0

3 years ago

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

6 0

3 years ago

Calculate the [OH-]<br> [H+]=0.000025

Harman [31]

Explanation:

There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.

3 0

3 years ago

For 30 points! Please help me understand this question

oksian1 [2.3K]

Answer:

D) Adding a catalyst

Explanation:

Adding a catalyst decreases activation energy and allows the reaction to occur more easily.

7 0

3 years ago

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

Marianna [84]

Answer:

0.136g

Explanation:

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

Initial mole of Co(NO3)2 $=\frac{mass}{molar mass}$

$=\frac{5.00}{182.94} \\\\=0.02733mol$

Mole of Co(NO3)2 in final solution

$=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol$

Mole of NO3- in final solution = 2 x Mole of Co(NO3)2

$=2\times 0.001093\\\\=0.002186mol$

Mass of NO3- in final solution is mole x Molar mass of NO3

$=0.002186\times62.01\\\\=0.136g$

6 0

3 years ago