All categories

2 years ago

Which of the following is an example of the law of multiple proportions?

1 answer:

3 0

The answer option which is an example of the law of multiple proportions is: B) Two different compounds formed from carbon and oxygen have the following mass ratios: 1.33 g O: 1 g C and 2.66 g O: 1 g C.

The Law of Multiple Proportions is also referred to as Dalton's Law and it states that when two chemical elements combine to form more than one chemical compound, the masses (weights) of one chemical element that combine with a fixed mass (weight) of the other chemical element will always be in a ratio of small whole numbers.

According to the Law of Multiple Proportions, when two chemical elements combine to form two different chemical compound, the masses (weights) of one chemical element that combine with 1 gram of the other chemical element can be expressed as a ratio of small whole numbers.

For example, carbon and oxygen react to form either carbon monoxide and carbon dioxide.

Carbon monoxide (CO):

12 grams of C = 16 grams of O.

1 gram of C = 1.33 gram of O.

Carbon dioxide ( $CO_2$ ):

12 gram of C = 32 grams of O.

1 gram of C = 2.66 gram of O.

Ratio of oxygen (O) = 16:32 = 1:2 (Law of Multiple Proportions).

Your question is lacking the necessary answer options, so I will be adding them here:

A. A sample of chlorine is found to contain three times as much Cl-35 as Cl-37.

B. Two different compounds formed from carbon and oxygen have the following mass ratios: 1.33 g O: 1 g C and 2.66 g O: 1 g C.

C. Two different samples of table salt are found to have the same ratio of sodium to chlorine.

D. The atomic mass of bromine is found to be 79.90 amu.

E. Nitrogen dioxide always has a mass ratio of 2.28 g O: 1 g N.

You might be interested in

A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

3 0

3 years ago

If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?

lawyer [7]

Answer:

${ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance = \frac{50}{ {4.5}^{2} } }} \\ \\ { \rm{resistance = 2.5 \: ohms}}$

8 0

2 years ago

RANGKAIAN RLC SERI PADA TEGANGAN BOLAK BALIK 220V R=80 OHM XL= 90 OHM XC = 30 OHM MAKA TEGANGAN PADDA INDUKTOR ADALAH???

Whitepunk [10]

Okay, 90% of this is nonsense besides the numbers maybe.

8 0

3 years ago

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

3 years ago

Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium

ivolga24 [154]

Answer:

Explanation:

Iron and gadlinium are both very easily made into magnetic substances. Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

5 0

3 years ago