Which of these is a common to all chemical changes

Find the number of neutron in an atom represented by 45X21

Helga [31]

Answer:

My school has no stationary and my teacher has no stationary

Explanation:

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5 0

2 years ago

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

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.

azamat

Answer:

(a) $m_{gold}=7.322g$

(b)

$V_{gold}=0.379cm^3$

$V_{copper}=0.122cm^3$

(c) $\rho _{coin}=15.94g/cm^3$

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

$m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g$

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

$V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3$