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Leya [2.2K]
3 years ago
9

How long is a wave traveling at 120 m/s if it has a frequency of 30 Hz?

Chemistry
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

4 m

Explanation:

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One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 ℃, and the temperature af
gtnhenbr [62]

Answer:fH = - 3,255.7 kJ/mol

Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0
3 years ago
Y Ggg I Ip us Edd u I by
Dimas [21]

Answer:

??

Explanation:

3 0
3 years ago
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Aquafers are bodies of water found above ground with thriving ecosystems.<br> O True<br> O False
BaLLatris [955]

Answer:

true

Explanation:

8 0
2 years ago
If D+2 would react with E!, what do you predict to be the formula?
kherson [118]

Answer:

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Explanation:

6 0
2 years ago
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
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