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vichka [17]
4 years ago
15

The colorless, oily, addictive compound found in cigarette smoke is ____.​

Chemistry
1 answer:
Minchanka [31]4 years ago
4 0
The answer is Nicotine.

Hope that helps!
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The enegry need to pump sodium outside the cell in active transport is
tester [92]
Adenosine triposphate (aka ATP)
5 0
4 years ago
From data below, calculate the total heat (in J) associated with the conversion of 0.499 mol ethanol gas (C2H6O) at 301°C and 1
8_murik_8 [283]

Answer:

The total heat associated is -30,520.3 J.

Explanation:

Moles of ethanol = 0.499 moles

Molar mass of ethanol = 46 g/mol  

Mass = Moles × Molar mass = 0.499 moles × 46 g/mol = 22.954 g

m is the mass of ethanol = 22.954 g

Q₁ is heat involved in the conversion of 22.954 g ethanol gas at 301°C to ethanol gas at 78.5°C

Thus, Q₁ = m × c × ΔT

Where,  

c =  The specific heat of the gas = 1.43 J/g°C

ΔT = Final temperature - Initial temperature = 78.5 - 301°C  

= - 222.5 °C

Applying the values in the above equation as:-

Q_1 = 22.954 g\times 1.43 J/g^0C\times (-222.5^oC) = -7303.3J

Q₂ is the enthalpy of condensation from gas to liquid for the given mass of ethanol .

Thus, Q₂ = moles×ΔH condensation  

Given that:- ΔH vaporization = 40.5 kJ/mol

Enthalpy of condensation of gaseous ethanol to liquid ethanol = - 40.5 kJ/mol

Considering, 1 kJ = 1000 J

So,  

ΔH condensation = - 40.5 ×1000 J/mol = - 40500 J/mol

Thus, Q₂ = 0.499 moles × (- 40500 J/mol) = -20209.5 J

Q₃ is heat involved in the conversion of 22.954 g gaseous ethanol at 78.5°C to ethanol liquid at 25.0°C.

Thus, Q₃ = m × C ×ΔT

Where,  

C = The specific heat of the liquid = 2.45 J/g°C

ΔT = Final temperature - Initial temperature = 25.0 - 78.5 °C  

= - 53.5 °C

Applying the values in the above equation as:-

Q_3 = 22.954 g\times 2.45 J/g^0C\times (-53.5 ^0C)=-3007.5 J

Applying the values as:

Total heat = Q_1+Q_2+Q_3

= -7303.3 J - 20209.5 J - 3007.5 J

= -30,520.3 J

The total heat associated is -30,520.3 J.

4 0
4 years ago
Please help, I’ll mark your answer as brainliest
Sedaia [141]

Answer:

5.16 gm of SO3 formed with 2 g of S

Explanation:

Mole weight of  S in the equation = 2 * 32 = 62 gm

Mole weight os   O2  in the equation  6 * 16 =96 gm

From the BALANCED  equation   the grams of   S  to   O2  is

    62   to 96   so    2 g of  S will need  approx 3 gm of O2

            this shows that S is the limiting reactant------>

                            there will be O left over  (approx 1 gram)

SO3 mole weight produced from the equation is   2 (32)(3*16) = 160 gm

  62 gm of S produces 160 gm of SO3

   62/160 =  2 / x       x = 5.16 gm of  SO3   are formed

3 0
2 years ago
Which would be the best material to use for making tea kettles?
Wewaii [24]
It’s a subjective answer but glass is believed to be the most safe material to use for tea kettles
6 0
3 years ago
What Happens When extra Solute is added to unsaturated ,saturated and supersaturated solution​
Ymorist [56]

of water is 36.0 g. If any more NaCl is added past that point, it will not dissolve because the solution is saturated. If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated.

8 0
3 years ago
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