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3 years ago

Light waves travel much__________________________________ than sound waves.

Chemistry

1 answer:

LUCKY_DIMON [66]3 years ago

3 0

Yeah, a lot faster than sound.

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How many grams of oxygen would be needed to completely react with 18 grams of methane (CH4)?

S_A_V [24]

Answer:

35.91 grams of oxygen

Explanation:

6 0

3 years ago

Nitrogen monoxide and hydrogen react to form nitrogen and water, like this: 2NO+2 H2(g)-N29)+2H2O(9) Write the pressure equilibr

lara31 [8.8K]

Answer:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)² = Kp

Explanation:

The reaction is:

2NO + 2H₂ → N₂ + 2H₂O

The expression for Kp (pressure equilibrium constant) would be:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)²

There is another expression for Kp, where you work with Kc (equilibrium constant)

Kp = Kc (R.T)^Δn

where R is the Ideal Gases constant

T° is absolute temperature

Δn = moles of gases formed - moles of gases, I had initially

4 0

3 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

What is the Ionic Equation and Net Ionic Equation for BaCl2 + Al2(SO4)=> 3Ba(SO4) + 2AlCl3?

Nikitich [7]

I think you want to ask the complete ionic equation and net ionic equation. For complete ionic equation: 3Ba2+ + 6Cl- + 2Al3+ + 3SO42- -->3BaSO4 + 2Al3+ + 6Cl-. For net ionic equation: 3Ba2+ + 3SO42- -->3BaSO4.

8 0

3 years ago

Your assignment for this week is to create a game that can be played in the Olympics. The game you create can be a team sport or

ASHA 777 [7]

Answer:

Shshsjsjssjjsjs

Explanation:

Hhhsusiaiahshwhauwiwgcececececegeuwhwccwywusgwvwgshwuuwuev3c

Thank u. Hope u liked my answer

3 0

3 years ago