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3 years ago

Over-encouraging lesser skilled players can lead to embarrassment. a. True b. False

Chemistry

2 answers:

Alex Ar [27]3 years ago

7 0

Answer: Option (a) is correct.

Explanation:

A person with less skills needs to work more or practice more on his abilities in order to improve their athletic ability to build a stronger team.

Over encouragement to lesser skilled person can lead to the development of over confidence. As after over encouragement, the less skilled person can have the illusion of knowing each and everything which is not true for anyone.

As no one is perfect, therefore, over-encouraging lesser skilled players can lead to embarrassment.

Margarita [4]3 years ago

6 0

My answer would be a. True

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8. What is the mass of 4.50 x 1022 Cu atoms?

vladimir2022 [97]

Answer:

4.7485 g

Explanation:

4.50 x 10^22 Cu atoms * (1 mol Cu / 6.022 x 10^23 Cu atoms) * 63.546 g Cu/(mol Cu) = 4.7485 g

In every mole of Cu, there are 6.022 x 10^23 atoms (Avogadro's number). The molecular weight of copper is 63.546 g/mol.

5 0

2 years ago

What is the meaning of 3 CO2 in a chemical reaction?

puteri [66]

Three is simply the coefficient which tells you the amount of molecules. In this case, there are three carbon dioxide molecules. If you want to break it down further, it also means there is three carbon atoms and six oxygen atoms.

I hope this helps.

7 0

2 years ago

The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and

Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

Moles KSCN = Moles Ag⁺ in the filtrate:

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

Total moles Ag⁺:

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

Moles of Ag⁺ in the precipitate:

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

Moles AsO₄ = Moles As:

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

Moles As₂O₃:

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

Mass As₂O₃:

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

7 0

3 years ago

Given the following thermodynamic data, calculate the lattice energy of LiCl:

tiny-mole [99]

Answer:

$\boxed{\text{-862 kJ/mol}}$

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

E/kJ

(5) Li⁺(g) +e⁻ ⟶ Li(g) 520

(6) Li(g) ⟶ Li(s) -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409

(8) Cl(g) ⟶ ½Cl₂(g) -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻ +349

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862

The lattice energy of LiCl is $\boxed{\textbf{-862 kJ/mol}}$ .

3 0

2 years ago

What is the temperature shown on the thermometer below?

vaieri [72.5K]

Answer:

- 13 * c

Explanation:

hope it helps u.....

6 0

2 years ago