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mario62 [17]
3 years ago
11

What are some potential traits that are needed to become a super athlete?​

Chemistry
2 answers:
nekit [7.7K]3 years ago
7 0

\huge \boxed{ \fcolorbox{black}{pink}{Answer}}

<u>20 Distinguishing Personality Traits of High-Performing Athletes</u>

1. Self Confidence. “Self-Confidence” isn't just a phrase for cheesy motivational posters. ...

2. Strong Sense of Motivation. It takes more than a shiny medal or hefty check to motivate the world's best athletes. ...

3. Inner Desire to Succeed. ...

4. Natural Goal Setter. ...

5. Self-Discipline. ...

6. Optimism. ...

7. Sense of Belonging. ...

8. Natural Leader.

konstantin123 [22]3 years ago
5 0

Answer: Confidence, strength, potential

Explanation:

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Morphine is a well known pain killer but is highly addictive. The lethal dose of morphine varies from person to person based on
Aliun [14]

Answer:

0.252 milimoles

Explanation:

To convert mass of a substance to moles it is necessary to use the molar mass of the substance.

The formula of morphine is C₁₇H₁₉NO₃, thus, its molar mass is:

C: 17*12.01g/mol = 204.17g/mol

H: 19*1.01g/mol = 19.19g/mol

N: 1*14g/mol = 14g/mol

O: 3*16g/mol = 48g/mol.

204.17 + 19.19 + 14 + 16 = <em>285.36g/mol</em>

Thus, moles of 71.891 mg = 0.071891g:

0.071891g × (1mol / 285.36g) = 2.5193x10⁻⁴ moles

As 1 mole = 1000 milimoles:

2.5193x10⁻⁴ moles = <em>0.252 milimoles</em>

7 0
3 years ago
A force interaction requires at least a(n)
Anastaziya [24]
C.) Action force unless it has potential/stored enerygy

8 0
3 years ago
Read 2 more answers
This table has information about the heat of fusion and the heat of vaporization of different substances. A 3-column table with
blagie [28]

H₂S

<h3>Further explanation</h3>

Given

ΔH fusion and ΔH vaporization  of different substances

Required

The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes

Solution

We can use the formula :

\tt \Delta H=\dfrac{Q}{n}

Q=heat/energy absorbed

n = moles

The heat absorbed : 58.16 kJ

moles = 3.11

so ΔH vaporization :

\tt \Delta H_{vap}=\dfrac{58.16~kJ}{3.11~mol}\\\\\Delta H_{vap}=18.7~kJ/mol

The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S

(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)

3 0
3 years ago
Read 2 more answers
Consider the following chemical reaction in a small volume of solution. According to LeChatelier’s Principle, which perturbation
stiks02 [169]

Answer:

Decreasing the temperature will shift the equilibrium leftwards towards reactants.

Explanation:

Hello!

In this case, since the reaction between chromate anions and hydrogen ions yields dichromate anions, water and heat, we can infer this is an exothermic reaction by which heat is released (remember in endothermic reactions heat is absorbed as a reactant), it means that considering the LeChatelier’s which states that increasing the temperature of an exothermic reaction shifts the equilibrium leftwards since heat is a product, otherwise (decreasing the temperature) the equilibrium will be shifted rightwards.

Therefore, decreasing the temperature is the perturbation that will shift the equilibrium leftwards towards the reactants.

Best regards!

7 0
2 years ago
What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
7 0
3 years ago
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