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Lady_Fox [76]
2 years ago
6

Which event must always occur for a chemical reaction to occur

Chemistry
1 answer:
Lorico [155]2 years ago
8 0
There must be effective collisions between the reacting chemical particles in order for the chemical reaction to occur.
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Does 1 gallon or 100 gallons take longer to boil
Ilya [14]

Answer:

it takes 100 gallons longer to boil then it does 1 gallon

Explanation:

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4 0
2 years ago
"Similar to the moon, Mercury and Venus have phases<br> when seen from Earth."
kotegsom [21]

Answer:

In our solar system, Mercury and Venus are inferior planets: their orbits are entirely inside the Earth's orbit. When seen from the Earth, inferior planets go through phases, like the Moon's. An inferior planet on the same side of the sun as the Earth appears dark and is not easy to see.

Explanation:

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5 0
3 years ago
A sample of pure oxalic acid (H2C2O4.2H2O) weighs 0.2000 g and requires 30.12 ml of KOH solution for
ValentinkaMS [17]

 The molarity  of KOH  is  0.1055 M

 <u><em> calculation</em></u>

Step  1: write  the  equation  for reaction between H₂C₂O₄.2H₂O  and KOH

H₂C₂O₄.2H₂O  + 2 KOH   →    K₂C₂O₄ +4 H₂O

step 2: find the moles  of H₂C₂O₄.2H₂O

moles = mass÷ molar  mass

from  periodic  table the  molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4)  + 2(18)=126 g/mol

 = 0.2000 g ÷ 126 g/mol =0.00159  moles


step 3: use the  mole  ratio  to  calculate the moles of KOH

H₂C₂O₄.2H₂O : KOH  is 1:2

therefore the  moles of KOH  =0.00159 x 2 = 0.00318  moles

step 4: find molarity of KOH

molarity = moles/volume in liters

volume in liters = 30.12/1000=0.03012 L

molarity  is therefore = 0.00318/0.03012 =0.1055 M

5 0
3 years ago
One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expa
Verdich [7]

Answer:

a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K

b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K

c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J

d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K

Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

∴ n = 1 mole

∴ PV = RTn....ideal gas

∴ P1 = 10 atm

∴ R = 0.082 atm.L/K.mol

∴ T = 300 K = T2

∴ V2 = 3*V1

⇒ W = 0.....expands freely into vacuum

⇒ ΔU = Q = 0....first law

⇒ ΔS = -  nR Ln(P2/P1).....ideal gas

∴ V1*P1/T1 = V2*P2/T2

∴ T1 = T2 = 300 K

⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm

⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)

⇒ ΔS = 0.09 atm.L/K

∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0

b) heated reversibly at constant volume:

⇒ W = 0 ...at constant volume

∴ T2 = 400 K; T1 = 300 K

∴ V1 = V2

⇒ Q = ΔU = CvΔT....first law

∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas

∴ ΔT = 400 - 300 = 100 K

⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J

∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J

∴ ΔS = - nR Ln (P2/P1)

∴ P2/T2 = P1/T1...constant volume

∴ P1 = 3.33 atm

⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm

⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)

⇒ ΔS = - 0.0235 atm.L/K

c) reversibly expanded at constant temperature:

∴ T1 = T2 = 400K

∴ V2 = 3*V1

∴ ΔU = 0...constant temperature

⇒ Q = - W....fisrt law

∴ W = - ∫ PdV..... reversibly expansion

∴ P = nRT/V... ideal gas

⇒ W = - nRT ∫ dV/V

⇒ W = - nRT Ln (V2/V1)

⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)

⇒ W = - 9.134 J/K *400K = - 3653.545 J

⇒ Q = - W = 3653.545 J

⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0

∴ ΔS = - nR Ln(P2/P1)

∴ P1 = 4.44 atm

⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)

⇒ P2 = 4.44atm/3 = 1.48 atm

⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)

⇒ ΔS = -9.134J/K * 400K = - 3653.545 J

d) reversibly cooled at constant pressure:

∴ T2 = 300 K;  T1 = 400 K

∴ P2 = P1

⇒ Q = ΔH = CpΔT

∴ Cp = 20.8 J/K.mol

∴ ΔT = 300 - 400 = - 100 K

⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J

⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J

⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J

∴ ΔS = ∫ δQ/T = ∫ nCpdT/T

⇒ ΔS = nCp Ln (T2/T1)

⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K

7 0
3 years ago
Which substance is acting as the Brønsted-Lowry acid in the following chemical reaction? NH4 + OH- yields NH3 + H2O
Ket [755]
The <span> the Brønsted-Lowry acid donates H⁻.
In this reaction Particle that  loose H⁺ is A. NH4⁺ ion.</span>
6 0
3 years ago
Read 2 more answers
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