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tigry1 [53]
3 years ago
12

Un deportista de 72 kg trepa por una cuerda hasta una altura de 12m. Calcula el incremento de energía potencial gravitatoria que

ha experimentado.
Chemistry
1 answer:
Oxana [17]3 years ago
4 0

A 72 kg athlete climbs a rope to a height of 12m. Calculate the increase in gravitational potential energy it has experienced.

Answer:

8467.2J

Explanation:

Given parameters:

Mass of the athlete = 72kg

Height of the climb  = 12m

Unknown:

Increase in gravitational potential energy it has experienced = ?

Solution:

Gravitational potential energy is the energy due to the position of a body. It is mathematically expressed as;

  Gravitational potential energy  = m x g x h

m is the mass

g is the acceleration due to gravity  = 9.8m/s²

h is the height

   Insert the parameters and solve;

  Gravitational potential energy  = 72 x 9.8 x 12

  GPE = 8467.2J

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Describe the contributions made by the following scientists to the development of the periodic table.
Basile [38]

Answer:

B

Explanation:

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6 0
3 years ago
Which set of coefficients will balance this chemical equation?
Montano1993 [528]

Answer:

Option B is correct = 1,3

Explanation:

Chemical equation:

C₂H₄ + O₂     →    CO₂ + H₂O

Balanced chemical equation:

C₂H₄ + 3O₂     →    2CO₂ + 2H₂O

Step 1:

Left side                      Right side

C = 2                           C = 1

H = 4                           O = 3

O = 2                           H = 2

Step 2:

C₂H₄ + O₂     →    2CO₂ + H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 5

O = 2                           H = 2

Step 3:

C₂H₄ + O₂     →    2CO₂ + 2H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 6

O = 2                           H = 4

Step 4:

C₂H₄ + 3O₂     →    2CO₂ + 2H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 6

O = 6                           H = 4

5 0
3 years ago
Grignard reactions are usually performed in ether solvent. Which statement below does not identify a safety concern associated w
nikdorinn [45]

Answer: A.  Diethyl ether has a very low miscibility in wate

The fact that the diethyl ether is miscible or not in water <u>does not imply a ris</u>k for the person who is working with this reagent in the laboratory.

However, the fact that diethyl ether forms explosive peroxides and that it is highly flammable implies that there is a risk of explosion when exposed to air and sunlight. On the other hand, as option C mentions, if a person inhales a large quantity of this reagent, they may lose consciousness and suffer some injury when fainting, due to the powerful anesthetic effect of this reagent.<u> In conclusion, options B, C and D are statements that imply safety problems associated with the use of diethyl ether in the laboratory, while option A does not imply it.</u>

6 0
3 years ago
Determine the empirical formulas for compounds with the following percent compositions:
Elis [28]

<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is P_2O_5

<u>For b:</u> The empirical formula of the compound is KH_2PO_4

<u>Explanation:</u>

  • <u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = \frac{1.406}{1.406}=1

For Oxygen = \frac{3.525}{1.406}=2.5

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = 1\times 2=2

For Oxygen = 2.5\times 2=5

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is P_2O_5

  • <u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium =\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = \frac{0.736}{0.735}=1

For Hydrogen = \frac{1.5}{0.735}=2.04\approx 2

For Phosphorus = \frac{0.735}{0.735}=1

For Oxygen = \frac{2.9375}{0.735}=3.99\approx 4

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is KH_2PO_4

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