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3 years ago

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What is the weight of a 90kg man standing on the moon, where gmoon = 1.64 m/s2?

1 answer:

Doss [256]3 years ago

3 0

Answer:

The weigth of a 90kg man standing on the moon is <u><em>147.6 N (option C)</em></u>

Explanation:

Weight is called the action exerted by the force of gravity on the body.

The mass (amount of matter that a body contains) of an object will always be the same, regardless of where it is located. Instead, the weight of the object will vary according to the force of gravity acting on it.

The formula that allows you to calculate the weight of any body is:

W = m*g

where:

W = weight measured in N.
m = mass measured in kg.
g = acceleration of gravity measured in m/s². The acceleration of gravity g is the same for all objects that fall due to gravitational attraction, whatever their size or composition. For example, as an approximate value on Earth, g = 9.8 m/s².

In this case, the mass m has a value of 90 kg and the gravity g has a value of 1.64 m/s², which is the value of the acceleration of gravity of the moon. Then:

W=90 kg* 1.64 m/s²

<u><em>W= 147.6 N</em></u>

Finally, <u><em>the weigth of a 90kg man standing on the moon is 147.6 N (option C)</em></u>

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Compare the amount of heat required to vaporize a 200.-gram sample of H2O(ℓ) at its boiling point to the amount of heat required

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A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

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<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

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