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3 years ago

PLEASE HELP ME! What are some patterns or trends that are present in the table of elements?

Chemistry

1 answer:

Sati [7]3 years ago

5 0

Some patterns and trend that are present in the periodic table would be

1. electronegativity (from left-to-right it increases across the table)

2. ionization (from left-to right it increases and from bottom-to-top it increases)

3. electron affinity (same as ionization energy)

4. atom radius (increases opposite way; from right-to-left it increases and from top-to-bottom it increases)

5. melting point (higher melting points with metals and lower melting point with non-metals)

6. metallic character (same as atom radius)

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Why alkali metals easly ionize

Dmitriy789 [7]

Here’s what I found:

It takes very little energy to remove that outermost electron from an alkali metal. Thus, alkali metals easily lose their outermost electron to become a +1 ion. ... In fact, as you go down the 1A column, the first ionization energies get lower and lower, making cesium the most easily ionized element on the periodic table.

So basically it’s because part of what makes alkali metals so reactive is that they have one electron in their outermost electron layer.

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3 years ago

Define the following terms<br><br> Uniformitarianism<br><br> & <br><br> trace fossil

AnnZ [28]

Answer:

Uniformitarianism- A theory that geologic process that occur today are similar to those that have occurred in the past

Trace Fossil- A fossil of a foot or trail or other travel of an animal rather than of the animal itself

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3 years ago

Read 2 more answers

A student needs to prepare 50.0 mL of a 1.30 M aqueous H2O2 solution. Calculate the volume of 5.0 M H2O2 stock solution that sho

lisov135 [29]

Answer:

13.0mL

Explanation:

So this is a classic M1V1=M2V2 problem, where M is molarity and V is volume and the subsequent numbers represent the two sets of condition (1 being before dilution and 2 being after dilution)

So M1 is going to be 5.0M because it is our initial molarity and V1 is what we are trying to find since we are trying to find how much of initial volume should be diluted.

M2 is 1.30M since it is what molarity is after dilution and M2 is what volume is after dilution which is 50mL.

So M1V1=M2V2 (becomes an algebra problem)

5*V1=1.30*50

V1= 13.0mL

Now this answer should make sense since to dilute something with large molarity to small you only need very few mL than the final volume because you add water to dilute it.

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3 years ago

Which square (A, B, C, or D) represents the repeating subunit of the polymer shown?

Lorico [155]

Answer:

Explanation:

That is because the monomer is a complete unit that repeats itself.

5 0

2 years ago

2. A sample of a gas is occupying a 1500 mL container at a pressure of 3.4 atm and a temperature of 25

lora16 [44]

2. The new pressure, given the data is 3.0 atm

3. The new temperature in K is 361 K

4. The new temperature in K is 348 K

<h3>2. How to determine the new pressure</h3>

Initial volume (V₁) = 1500 mL
Initial pressure (P₁) = 3.4 atm
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
New temperature (T₂) = 75 °C = 75 + 273 = 348 K
New Volume (V₂) = 2000 mL
New pressure (P₂) = ?

The new pressure of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

(3.4 × 1500) / 298 = (P₂ × 2000) / 348

Cross multiply

P₂ × 2000 × 298 = 3.4 × 1500 × 348

Divide both sides by 2000 × 298

P₂ = (3.4 × 1500 × 348) / (2000 × 298)

P₂ = 3.0 atm

<h3>3. How to determine the new temperature</h3>

Initial volume (V₁) = 450 mL
Initial pressure (P₁) = 167 KPa
Initial temperature (T₁) = 295 K
New pressure (P₂) = 230 KPa
New Volume (V₂) = 400 mL
New temperature (T₂) =?

The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

(167 × 450) / 295 = (230 × 400) / T₂

Cross multiply

T₂ × 167 × 450 = 295 × 230 × 400

Divide both sides by 167 × 450

T₂ = (295 × 230 × 400) / (167 × 450)

T₂ = 361 K

<h3>4. How to determine the new temperature</h3>

Initial volume (V₁) = 3.6 L
Initial pressure (P₁) = 9.2 atm
Initial temperature (T₁) = 298 K
New Volume (V₂) = 5.3 L
New pressure (P₂) = 7.3 atm
New temperature (T₂) =?

The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

(9.2 × 3.6) / 298 = (7.3 × 5.3) / T₂

Cross multiply

T₂ × 9.2 × 3.6 = 298 × 7.3 × 5.3

Divide both sides by 9.2 × 3.6

T₂ = (298 × 7.3 × 5.3) / (9.2 × 3.6)

T₂ = 348 K

Learn more about gas laws:

brainly.com/question/6844441

#SPJ1

7 0

2 years ago