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gtnhenbr [62]
3 years ago
9

Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dyne

s of a 15-ton object (not metric tons)
Chemistry
1 answer:
belka [17]3 years ago
4 0

Answer:

a) 965,1 lbf

b) 4,5 kg

c) 1,33 * 10^6 dynes

Explanation:

Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.

Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.

                                        w=mg

In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International  System) or 32,17 ft/s² (in the FPS system).

To solve this problem we'll use the following conversion factors:

1 lbf = 1 lbm*ft/s²

1 N = 1 kg*m/s²

1 dyne = 1 gr*cm/s²   and 1 N =10^5 dynes

1 ton = 907,18 kg

1 k = 1000 gr

a) m = 30 lbm

w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf

b) w = 44 N

First, we clear m of the weight equation and then we replace our data.

m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg

c) m = 15 ton

m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes

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An unknown compound contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 2.77 g sample of the compound
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The molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

<h3>Molecular formula of the compound</h3>

The molecular formula is calculated as follows;

CHNO  +  O₂ ------------> CO₂  + H₂O

Mass of carbon, C:  = (6.64 x 12)/44 = 1.81 g in 2.77 g sample

Mass of hydrogen, H: = (1.67 x 2)/18 = 0.186 g in 2.77 g sample

Mass of Nitrogen, N: = (2.77 x 0.143)/1.69 = 0.234 g

Mass of oxygen, O:  = 2.77 g - 1.81 g - 0.186 g - 0.234 g = 0.54 g

<h3>molar ratio of the elements: </h3>

C = 1.81 g = 0.15 mol

H = 0.186 g = 0.186 mol

N = 0.234 g = 0.017 mol

O = 0.54 g = 0.0337 mol

divide through with the smallest number of moles (0.017 mol);

C = 9

H = 11

N = 1

O = 2

Molecular formula = C₉H₁₁NO₂

Check the molar mass of the compound = (9 x 12) + (11 x 1) + (14) + (2 x 16) = 165 g/mol

Thus, the molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

Learn more about molecular mass here: brainly.com/question/21334167

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An ethylene gas torch requires 300 L of gas at 0.8 atm. What will be the pressure of the gas if ethylene is supplied by a 200.0
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Answer:

1.2 atm

Explanation:

Given data

  • Volume of the gas in the tank (V₁): 200.0 L
  • Pressure of ethylene gas in the tank (P₁): ?
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If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.

P_1 \times V_1 = P_2 \times V_2\\P_1 = \frac{P_2 \times V_2}{V_1} = \frac{0.8atm \times 300L}{200.0L} = 1.2 atm

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