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kondor19780726 [428]
3 years ago
5

Given the unbalanced chemical equation below. What would be the coefficient of AgNO3?

Chemistry
1 answer:
OLga [1]3 years ago
8 0
2
I can’t really explain in words so I took a pic of the work I did (Ignore the worksheet and just look at what I wrote to balance the equation.)

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A partir de lo propuesto por Bohr en 1913, que postulado es correcto:
TEA [102]

Answer:

Primer postulado:

Así Bohr asumió que el átomo de hidrógeno puede existir solo en ciertos estados discretos, los cuales son denominados estados estacionarios del átomo. En el átomo no hay emisión de radiación electromagnética mientras el electrón no cambia de órbita.

Explanation:

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3 years ago
What were some of the “old” ideas that Isaac Newton learned in school?
Scilla [17]

Answer:

In addition to mathematics, physics and astronomy, Newton also had an interest in alchemy, mysticism, and theology.

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3 years ago
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Bronze is an alloy, which is what form of matter?
grandymaker [24]
It is a Compound. have a nice day
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3 years ago
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Cell notation will list each half-reaction:
likoan [24]

Answer:

on each side of the salt bridge, which is represented by a double vertical line

Explanation:

While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.

The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.

6 0
3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
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