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worty [1.4K]
3 years ago
5

Chemistry Grade 11: Hi, I don't know what this is, please help?

Chemistry
1 answer:
olga55 [171]3 years ago
4 0
So percent yield is actual yield/theoretical times 100. actual yield is the number a chemical produces. theoretical yield is what it should produce
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During a spontaneous chemical reaction, it is found that ΔSsys is less than 0. This means that Group of answer choices ΔSsurr is
Evgen [1.6K]

Answer:

The correct option is: ΔSsurr is greater than 0 and its magnitude is greater than ΔSsys.  

Explanation:

For a <u>chemical reaction to be spontaneous, the change in the</u> <u>Gibbs free energy should be negative (ΔG = -ve).</u>

At constant temperature (T) and pressure, the Gibbs free energy is given by the equation: \Delta G_{system} = \Delta H_{system} - T\Delta S_{system}

Here, \Delta H_{system} is the enthalpy change and \Delta S_{system} is the entropy change of the system.

Also, <u>according to the second law of thermodynamics</u>, <u>for a spontaneous process, the total entropy</u>, \Delta S_{universe} = \Delta S_{system} + \Delta S_{surrounding} > 0

Therefore, <u>for a reaction to be spontaneous</u>, if \Delta S_{system} is less than 0 or negative; then <u>\Delta S_{surrounding} should be greater than 0 or positive. </u>Also, \Delta S_{surrounding} > \Delta S_{system}.

3 0
3 years ago
Please help me!!!!!!!!
Oksanka [162]
Your answer would be surface runoff 
5 0
3 years ago
¿Por qué se generan rayos (tormenta eléctrica) cuando un volcán hace erupción ?
rosijanka [135]

Answer:

Los relámpagos volcánicos surgen de la colisión y fragmentación de partículas de ceniza volcánica (y, a veces, hielo), que generan electricidad estática dentro de la columna volcánica, lo que lleva al nombre de tormenta eléctrica sucia. La convección húmeda y la formación de hielo también impulsan la dinámica de la pluma de erupción y pueden desencadenar rayos volcánicos. por cierto utilicé un traductor espero que esto ayude ^^

Explanation:

5 0
3 years ago
2
gayaneshka [121]

Answer:

The answer is A. Key

Explanation:

4 0
3 years ago
How much work is required to lift a 0.500 kg block 18.5 m?
kipiarov [429]

Answer:

Work = 90.65 j

Explanation:

Given data:

Mass = 0.500 Kg

Distance = 18.5 m

Work done = ?

Solution:

Work = force . distance

Force = mg

Work = mg.distance

Work = mgh

Work = 0.500 Kg × 9.8 m/s²× 18.5 m

Work = 90.65 Kg .m²/s²

Kg .m²/s² = j

Work = 90.65 j

3 0
3 years ago
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