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Kitty [74]
3 years ago
6

Which of the following is the term used to describe a set of elements whose properties tend to be largely predictable based on p

osition on the periodic table?
Chemistry
2 answers:
lys-0071 [83]3 years ago
8 0
Group.... because elements in a group have same chemical properties
Airida [17]3 years ago
7 0

Explanation:

Elements which belong to same group show similar chemical properties.

This is because elements that lie in the same group have same number of valence electrons. Hence, they tend to show same type of reactivity which makes them exhibit similar chemical properties.

As chemical properties are defined as the properties which tend to show difference in chemical composition of a substance.

For example, toxicity, reactivity, combustion etc are all chemical properties.

Thus, we can conclude that group is the term used to describe a set of elements whose properties tend to be largely predictable based on position on the periodic table.

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How many molecules are in 41.8 g of sulfuric acid
Anton [14]

Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

3 0
1 year ago
What happens to the volume of a gas in a closed container if the temperature increases, but the pressure remains the same? Why?
labwork [276]

Answer:

Volume will goes to increase.

Explanation:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.

Now we will put the suppose values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 4.5 L × 348 K / 298 k

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

Hence prove that volume increase by increasing the temperature.

3 0
3 years ago
In each reaction box place the best reagent and conditions from the list below benzene 3 boxes
den301095 [7]

Answer : The correct answer is 1) AlCl₃ - CH₃Cl 2) HNO₃ -H₂SO₄ at room temperature 3) Fuming HNO₃ -H₂SO₄ at 90-100 ⁰ C heat .

I think this reaction is forming 2,4,6- trinitrotoluene from benzene, since the product is not mentioned. Following are the steps to convert Benzene to 2,4,6 trinitrotoluene .

Step 1: Conversion of Benzene to Toluene .

Benzene can be converted to toluene by Friedel Craft Alkylation of benzene . In this reaction reagent AlCl₃ and Ch3Cl is used . Electrophile CH³⁺ is produced which attached on carbon of benzene and formation of Toluene and HCl occur.

Benzene     \frac{AlCl3}{Ch3Cl}>   Toulene + HCl

Step 2 : Conversion of Toluene to dinitrotoluene.

Dinitritoluene is prepared from toluene by Nitration . This reaction uses Electrophilic substitution mechanism . The reagents used are HNO₃ and H₂SO₄ at room temperature . These reagents produces NO₂⁺ ( nitronium ion ), a electrophile which attacks on C2 and C4 Carbon atoms of Toluene.

Toluene Tolune   \frac{HNO3 -H2SO4}{30-40 degree C} ->  2,4- dinitrotoluene

Step 3) Conversion of Dinitro toluene to trinitrotoluene.

This reaction is extended nitration of toluene . Further nitration is done in extreme condition . The temperature of reaction is increased to 90- 100 ⁰ C . Due to which there is more production of NO²⁺ ion occurs from HNO₃ -H₂SO₄ and they attack on C6 carbon atom of dinitrotoluene which forms 2,4,6- trinitrotoluene.

Dinitrotoluene 2,4 -dinitrotoluene   \frac{fuming HNO3-H2So4}{90-100 C} ->  2,4,6-trinitrotoluene.

So over all reaction uses three reagents in order :

Benzene  \frac{AlCl3}{CH3Cl}  -> Toluene  \frac{HNO3-H2So4}{room temp}  -> 2,4-dinitrotoluene  \frac{Fuming HNO3 -H2SO4}{Heating at 90-100 C}  -> 2,4,6-trinitrotoluene .

3 0
4 years ago
Read 2 more answers
Give the n and l values and the number of orbitals for sublevel 5g.
Pepsi [2]

The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

There are total four quantum numbers:

1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4     ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1)  ( 2×4 + 1) = 9

Thus,  The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

brainly.com/question/14650894

#SPJ1

6 0
1 year ago
An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of
Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the  alloy is 2.324 J/g°C

3 0
3 years ago
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