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1 year ago

What circumstances during the French Revolution permitted the metric system to gain a foothold?

Chemistry

1 answer:

dem82 [27]1 year ago

8 0

Answer:

The trade was difficult with the old system.

Explanation:

The existing system for the measurement of goods did not work well in the trade because there are different system used by the people of different cities so metric system was used in the whole country in place of the old system of measurement during the French Revolution in order to make the trade easier and quick across the country. This metric system is extensively used in the trade as compared to the old system of measurement due to its easiness and fast.

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Why is it important to have regular supervision of the weight and measurements in the market

Sergio [31]

Answer:

Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.

Explanation:

4 0

1 year ago

MUSCLES IN A KANGAROO'S LEGS WORK BECAUSE OF THE CELLS THAT MAKE UP THE MUSCLE. WHICH COMPONENT OF CELL THEORY DOES THIS BEST IL

Fed [463]

Answer:

Explanation:

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1 year ago

Read 2 more answers

(a) What are the possible values of l for n = 4? (Enter your answers as a comma-separated list.)

vodka [1.7K]

Answer:

(a) 0,1,2,3 (b) -3,-2,-1,0,1,2,3 (c) 6 (d) 5

Explanation:

(a) for the principal quantum number 'n', the possible values of I = 0 to n-1. Thus, if the principal quantum number 'n' =4, I = 0,1,2,3.

(b) for a given number of 'I', the possible values of ml = -I to +I. Therefore, if I =3, then ml = -3,-2,-1,0,1,2,3

(c) 'I' which is the orbital angular momentum quantum number usually has values from 0,1,2,⋯,n−1. Therefore, for n greater than or 6, t would be greater than or equal to 5. Thus, the smallest possible value of n for which I can be 6 is 6.

(d) In a 3-dimensional figure, If the z-component of the orbital angular momentum Lz for which I=5 is measured, The possible outcomes will be:

mħ = -5ħ, -4ħ, -3ħ, -2ħ, -1ħ, 0, 1ħ, 2ħ, 3ħ, 4ħ, 5ħ.

Thus, the smallest possible l that can have a z component of 5ℏ is 5.

4 0

2 years ago

To 100.0 g water at 25.00 ºc in a well-insulated container is added a block of aluminum initially at 100.0 ºc. the temperature o

11111nata11111 [884]

When the amount of heat gained = the amount of heat loss

so, M*C*ΔTloses = M*C* ΔT gained

when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.

∴( M * C * ΔT )W = (M*C*ΔT) Al

when Mw is the mass of water = 100 g

and C the specific heat capacity of water = 4.18

and ΔT the change in temperature for water= 28-25 = 3 ° C

and ΔT the change in temperature for Al = 100-28= 72°C

and M Al is the mass of Al block

C is the specific heat capacity of the block = 0.9

so by substitution:

100 g * 4.18*3 = M Al * 0.9*72

∴ the mass of Al block is = 100 g *4.18 / 0.9*72

= 19.35 g

4 0

2 years ago

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y (1)

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol× $\frac{1mol}{80,043g}$ X

Y = 321J/g X (2)

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

X = 35,2g

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0

2 years ago