For X + Y → XY, if [XY] is increasing at a rate 0.10 M s-1, predict the rate of depletion of both reactants.

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

belka [17]

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

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

4 0

3 years ago

What is the mass of 2.80 grams of H2O

IrinaK [193]

The mass of 2.80 grams of h2o is 18.02 amu I believe

8 0

2 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

2 years ago

What<br>was the initial volume of the hydrogen in cm3?

svetlana [45]

Answer:

255.51cm3

Explanation:

Data obtained from the question include:

V1 (initial volume) =?

T1 (initial temperature) = 50°C = 50 + 273 = 323K

T2 (final temperature) = - 5°C = - 5 + 237 = 268K

V2 (final volume) = 212cm3

Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:

V1/T1 = V2/T2

V1/323 = 212/268

Cross multiply to express in linear form

V1 x 268 = 323 x 212

Divide both side by 268

V1 = (323 x 212)/268

V1 = 255.51cm3

Therefore, the initial volume of the gas is 255.51cm3

5 0

2 years ago

Read 2 more answers

Which component of a plant makes the cell walls around plant cells tough and allows branches, stems, and leaves to be strong

Vsevolod [243]

<h3><u>Answer;</u></h3>

Cellulose

<h3><u>Explanation</u>;</h3>

Cellulose is a polysaccharide and the most abundant organic compound on the Earth's surface.

<em><u>It is an important organic molecule due to its strong structure which provides a wide variety of functions. </u></em>
<em><u>Cellulose is a major component of tough cell walls that surround plant cells and is what makes plant stems, leaves, and branches very strong.</u></em>
The molecules of cellulose are arranged such that they are parallel to each other joined by hydrogen bond. this arrangement forms long structures that combine with other cellulose molecules producing a strong support structure.

8 0

2 years ago