Balsa wood has a typical density of 140kg/m^3 . What is the volume of a piece of balsa wood weighing 50lbs?

Answers

Answer 1

The volume of the balsa wood weighing 50lbs is approximately 0.162 cubic meters.

What is volume?

Volume is defined as the space occupied within the boundaries of an object in three-dimensional space.

To find the volume of a piece of balsa wood weighing 50lbs, we need to first convert the weight to kilograms and then divide by the density of balsa wood.

1 lb = 0.45359237 kg

So, 50 lbs = 50 * 0.45359237 = 22.6796185 kg

The volume of the balsa wood can then be calculated as follows:

Volume = 22.6796185 kg / (140 kg/m^3) = 0.162 m^3

Therefore, The volume of the balsa wood weighing 50lbs is approximately 0.162 cubic meters.

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Related Questions

The amount of energy required to change 20g of water to steam is +150 KJ/mol. Calculate the standard heat of vaporization of water.​

Answers

The amount of energy required to change 20g of water to steam is +150 KJ/mol. The standard heat of vaporization of water is 7.5 KJ/mol.

What is heat of vaporization ?

The amount of energy required to convert a liquid to a vapour at constant temperature and pressure is known as latent heat. The heat of fusion is the energy required to convert a solid to a liquid, and the heat of sublimation is the energy required to convert a solid to a gas.

Given:

q = 150 KJ/mol.

m = 20gram

The standard heat of vaporization of water is calculated by given formula

Hv = q / m

Where,

Hv is heat of vaporization of water

q is heat energy

m is mass

By substituting given value in equation we get,

Hv = 150 / 20

= 7.5 KJ/mol

Thus, the standard heat of vaporization of water is 7.5 KJ/mol.

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118Sn 117.90 47.%
119Sn 118.90 53.%

Use these measurements to complete the entry for tin in the Periodic Table that would be used on 98765 ALEKS. Round your entry for the atomic mass to 3 significant digits.

___
Sn
___

Answers

The average atomic mass of tin is 118.

What is the relative atomic mass of tin?

The average atomic mass of an element can be calculated by taking into account the isotopic composition of the element and the atomic masses of its isotopes. The average atomic mass is found by weighting the atomic masses of each isotope with its relative abundance, and then summing these values.

Average atomic mass = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2) + ... + (mass of isotope n * abundance of isotope n)

For the average atomic mas of tin;

(117.9 * 0.47) + (118.9 * 0.53) =

55.4 + 63.0

= 118.4

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A reaction has a rate constant of 0.0017/s at 400.0 K and 0.689/s at 450.0 K.
a. Determine the activation barrier for the reaction.
b. What is the value of the rate constant at 425 K?

Answers

The molecules mix more slowly in a low-rate reaction than they do in an elevated reaction. Therefore, the activation barrier for the reaction is 122 kJ.

What is rate of reaction?

The change in concentration of each of the products or reactants each unit of time is referred to as the rate of reaction. The pace that a chemical reaction takes place is known as the reaction rate. The molecules mix more slowly in a low-rate reaction than they do in an elevated reaction.

log(k[tex]_2[/tex] / k[tex]_1[/tex])=(-Ea/R)×(1/T[tex]_1[/tex]-1/T[tex]_2[/tex])

Where,

k[tex]_1[/tex] = 0.0117 s-1

k[tex]_2[/tex]= 0.689 s-1

T[tex]_1[/tex] = 400.0 k

T[tex]_2[/tex] = 450.0 k

R is the ideal gas constant

R = 8.314 KJ / (Kmol * K)

Substituting all the given values, we get

ln (0.0117/0.689)=-Ea/(8.314)*((1/400)-(1/450))

Ea = 122 kJ

Therefore, the activation barrier for the reaction is 122 kJ.

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HELPPP PLEASEEEEE!!!!!!!

Answers

The evidence in figure 2 and table 1 which describes the changes in the rock after a year is the presence of a round edge.

What is Weathering?

This is referred to as the breaking down or dissolving of rocks and minerals on Earths surface through elements such as wind, water etc.

In the diagram given, the rock previously had a sharp edge but as time went on there was breakdown of the edges which resulted in it becoming round and is therefore the evidence that weathering most likely occurred in the rock.

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A sample of trifluoroacetic acid, C2HF3O2, contains 47.3 g of oxygen. Calculate the mass of the trifluoroacetic acid sample.

Answers

Based on the percentage mass of oxygen in trifluoroacetic acid, the mass of the trifluoroacetic acid sample is 168.5 grams.

What is the percentage mass of oxygen in trifluoroacetic acid?

The percentage mass of oxygen in trifluoroacetic acid is determined as follows:

The molar mass of trifluoroacetic acid = 12 * 2 + 1 + 19 * 3 + 16 * 2

The molar mass of trifluoroacetic acid = 114 g

The percentage mass of oxygen in trifluoroacetic acid = (16 * 2)/114 * 100%

The percentage mass of oxygen in trifluoroacetic acid = 28.07%

The mass of the trifluoroacetic acid sample will be 47.3/ 28.07% = 168.5 grams

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The pressure of a gas sample at 37.0 °C is increased from 1.10 atm to 2.30 atm by heating at a constant volume. What is the temperature (in °C) of the gas at 2.30 atm?

Answers

Considering the Gay Lussac's Law, the the temperature of the gas at 2.30 atm is 375.18 °C.

Definition of Gay Lussac's Law

Gay Lussac's Law establishes the relationship between the pressure and temperature of a gas when the volume is constant.

This law establishes that the pressure of the gas is directly proportional to its temperature because as the temperature increases the gas molecules move faster and therefore the number of collisions against the walls increases. In other words, the pressure increases since the container has fixed walls.

Mathematically, Gay Lussac's Law says that the ratio between pressure and temperature is constant:

P÷T=k

where

P is the pressure.T is the temperature.k is a constant.

Analyzing an initial state 1 and a final state 2, it is fulfilled:

P₁÷T₁=P₂÷T₂

Temperature at 2.30 atm

In this case, you know:

P₁= 1.10 atmT₁= 37 °C= 310 K (being 0°C= 273 K)P₂= 2.30 atm T₂= ?

Replacing in Gay Lussac's law:

1.10 atm÷ 310 K= 2.30 atm÷T₂

Solving:

(1.10 atm÷ 310 K)×T₂= 2.30 atm

T₂= 2.30 atm÷ (1.10 atm÷ 310 K)

T₂= 648.18 K= 375.18 °C

Finally, the final temperature is 375.18 °C.

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