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ZanzabumX [31]
3 years ago
9

A piece of hot metal is put into an equal mass of water. The water temperature rises 5o and the metal’s temperature drops 10o. W

hat is the specific heat of the metal?
Chemistry
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

The answer to your question is     Cm = 2.093 J/kg°C

Explanation:

Data

mass metal = mass of water

Δ water = 5°

Δ metal = 10°

Cm = ?

Cw = 4.186 J/Kg°C

Formula

              for metal       for water

               mCmΔTm  = mCwΔTw

Cancel mass of metal

                  Cm ΔTm  =  CwΔTw

solve for Cm

                         Cm = (CwΔTw) / ΔTm

Substitution

                         Cm = (4.186 x 5) / 10

Simplification

                         Cm = 20.93 / 10

Result

                         Cm = 2.093 J/kg°C

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84.99467

Explanation:

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the density of dry air at 20 degrees celsius is 1.20 g/L. What is the mass of air, in kilograms, of a room that measures 24.0m b
nika2105 [10]

Answer:

1728 kg is the mass of air for the room

Explanation:

An exercise of unit conversion.

The volume of the room will be:

24 m . 15 m . 4m = 1440 m³

Density of air is 1.20 g/L which means that in 1L, there is contained 1.20 g of air.

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1440 m³ . 1000 = 1440000 dm³ ⇒ 1.44x10⁶ L

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1.20 g/L = mass of air / 1.44x10⁶ L

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7 0
3 years ago
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion,
Norma-Jean [14]

Answer:

m_{NaNH_2}=30.42gNaNH_2

m_{H_2}=0.783gH_2

Explanation:

Hello,

In this case, the reaction between sodium and ammonia is:

2Na+2NH_3\rightarrow 2NaNH_2+H_2

Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:

n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa

And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):

n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa

In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:

m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2

m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2

Best regards.

3 0
3 years ago
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Answer:

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