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3 years ago

I WILL GIVE BRAINLIEST!!!!

Chemistry

2 answers:

Alona [7]3 years ago

7 0

A. Atomic radius increases

Harlamova29_29 [7]3 years ago

3 0

It’s B. Atomic radius decreases from left to right

Hope that I could help you

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Which of the following is a value of biodiversity?

Sever21 [200]

Biodiversity has a fundamental value to humans because we are so dependent on it for our cultural, economic, and environmental well-being. Some argue that it is our moral responsibility to preserve the Earth’s incredible diversity for the next generation. Others simply like knowing that nature’s great diversity exists and that the opportunity to utilize it later, if need be, is secure. Scientists value biodiversity because it offers clues about natural systems that we are still trying to understand. Arguably, the greatest value to humans, however, comes from the ?ecosystem services? it provides.

Biodiversity forms the backbone of viable ecosystems on which we depend on for basic necessities, security, and health. By breaking down plant and animal matter, for example, insects and other invertebrates make nutrients available to plants and are integral to the carbon and nitrogen cycles. Other species pollinate crops, an essential service for farmers. Healthy ecosystems can mitigate or prevent flooding, erosion, and other natural disasters. These ecosystem services also play a hand in the functioning of our climate and in both air and water quality.

4 0

3 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

4 years ago

Problem PageQuestion A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of

igomit [66]

Answer:

5.37 × 10⁻⁴ mol/L

Explanation:

<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>

Step 1: Given data

Initial concentration (C₁): 0.00154 mol/L

Initial volume (V₁): 230. mL

Final concentration (C₂): ?

Final volume (V₂): 660. mL

Step 2: Calculate the concentration of the final solution

We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L

5 0

3 years ago

How can you increase friction

Lelu [443]

Answer:Well There is many ways to increase friction between objects. One being rubbing the two objects together quicker and harder. If there is any sort of wetness or anything related to that make sure to dry the surface between the two objects you want to create friction between so it will be more effective.

Explanation:

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3 years ago

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. In which reaction is nitric acid acting as an oxidising agent?

Talja [164]

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.

Regards.

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3 years ago

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