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Elan Coil [88]
3 years ago
8

which of the following actions can result in decreasing the pressure of a gas inside a sealed container: increasing the collisio

n rate between particles, increasing the speed at which the gas particles move, decreasing the collision rate between particles or decreasing the speed at which the gas particles move
Chemistry
1 answer:
nekit [7.7K]3 years ago
4 0
Decreasing the speed at which particles move. 
That is the answer because less speed involves less energy to move each particle which translates to less energy being exerted on the container when the particle and the container collide. That would result in less pressure.
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REVERSE REACTION WILL BE SO2+O2
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A police car is approaching the corner. Explain what happens to the frequency of sound as it draws closer. What will you hear?
barxatty [35]
The frequency stays the same it just gets louder
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Complete this neutralization equation:<br> H2SO4 + Al(OH)3
elena-s [515]

Answer:

H2SO4 + Al(OH)3 = Al2(SO4)3 + H2O

Explanation:

4 0
4 years ago
Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----&gt; 2Cr3+ + 3Cl2 + 7H2O The reac
stich3 [128]

Answer:

Value of Q_{p} for the given redox reaction is 1.0\times 10^{-8}

Explanation:

Redox reaction with states of species:

14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)

Reaction quotient for this redox reaction:

Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of H_{2}O is taken as 1 due to the fact that H_{2}O is a pure liquid.

pH=-log[H^{+}]

So, [H^{+}]=10^{-pH}

Plug in all the given values in the equation of Q_{p}:

Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}

7 0
3 years ago
Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp
Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

         Equilibrium when T = 100 °C

      Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must <em>add heat</em> to boil water

ΔS > 0 (positive), because changing from a liquid to a gas i<em>ncreases the disorder </em>

ΔG = 0, because the liquid-vapour equilibrium process is at <em>equilibrium</em> at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.


If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.


If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0
4 years ago
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