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irina1246 [14]
2 years ago
11

Calculate the number of moles when 3.00l of a gas at 814 mmhg and 25 °c?

Chemistry
1 answer:
Mariulka [41]2 years ago
5 0
  • Volume=V=3L
  • Pressure=P=814mmHg
  • Temp=T=25°C

So

No of moles=n

\\ \rm\rightarrowtail PV=nRT

\\ \rm\rightarrowtail n=\dfrac{PV}{RT}

\\ \rm\rightarrowtail n=\dfrac{814(3)}{25R}

\\ \rm\rightarrowtail n=2442/25R

\\ \rm\rightarrowtail n=97.7Rmol

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Most indoor growers will want to aim for 50-70% relative humidity, although 60-80% is usually tolerable.

Explanation:

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Urgent!!!
Yakvenalex [24]

The specific heat of the metal, given the data from the question is 0.60 J/gºC

<h3>Data obtained from the question </h3>

The following data were obtained from the question:

  • Mass of metal (M) = 74 g
  • Temperature of metal (T) = 94 °C
  • Mass of water (Mᵥᵥ) = 120 g
  • Temperature of water (Tᵥᵥ) = 26.5 °C
  • Equilibrium temperature (Tₑ) = 32 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Specific heat capacity of metal (C) =?

<h3>How to determine the specific heat capacity of the metal</h3>

The specific heat capacity of the sample of gold can be obtained as follow:

According to the law of conservation of energy, we have:

Heat loss = Heat gain

MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)

74 × C(94 – 32) = 120 × 4.184 (32 – 26.5)

C × 4588 = 2761.44

Divide both side by 4588

C = 2761.44 / 4588

C = 0.60 J/gºC

Thus, the specific heat capacity of the metal is 0.60 J/gºC

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

6 0
1 year ago
How<br> much heat must be added to<br> 85g<br> of water<br> to raise its temperature 225?
ki77a [65]

Answer:

One of water's most significant properties is that it takes a lot of heat to it to make it get hot. Precisely, water has to absorb 4,184 Joules of heat (1 calorie) for the temperature of one kilogram of water to increase 1°C.

Explanation:

7 0
3 years ago
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c
gtnhenbr [62]

Answer:

1.13 × 10⁶ g

Explanation:

Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.

Al³⁺ + 3 e⁻ → Al

We can establish the following relations:

  • 1 Ampere = 1 Coulomb / second
  • The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
  • 1 mole of Al is produced when 3 moles of electrons circulate
  • The molar mass of Al is 26.98 g/mol.

The mass of aluminum produced under these conditions is:

90.0 s \times \frac{1s}{620c} \times \frac{96,468c}{1mole^{-} } \times \frac{3mole^{-}}{1molAl} \times \frac{26.98gAl}{1molAl} =1.13 \times 10^{6} g Al

6 0
4 years ago
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