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3 years ago

¿En qué fecha, Mendeléiev creo su tabla periódica según la leyenda?

Chemistry

1 answer:

musickatia [10]3 years ago

6 0

Answer:

Legend has it that Mendeleev conceived and created his table in a single day: February 17, 1869, on the Russian calendar (March 1 in most of the rest of the world).READ THE EXPLANATION

Explanation:

La leyenda dice que Mendeleev concibió y creó su mesa en un solo día: 17 de febrero de 1869, en el calendario ruso (1 de marzo en la mayoría del resto del mundo).

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Which element would be a positive ion in a compound: Sr or Te

tamaranim1 [39]

I'm certain that the answer is Sr

5 0

3 years ago

A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder

Mazyrski [523]

Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

$PV_1=nRT$

where,

P = pressure of gas = 1 atm

$V_1$ = volume of gas = ?

T = temperature of gas = $270^oC=273+270=543K$

n = number of moles of gas = 0.205 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K$

$V_1=9.12L$

Now we have to calculate the volume at 24°C.

$PV_2=nRT$

where,

P = pressure of gas = 1 atm

$V_2$ = volume of gas = ?

T = temperature of gas = $24^oC=273+24=297K$

n = number of moles of gas = 0.205 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K$

$V_2=4.99L$

Now we have to calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 9.12 L

$V_2$ = final volume = 4.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (4.99-9.12)L$

$w=4.13L.artm=4.13\times 101.3J=418.4J$

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

6 0

3 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago

Can anyone answer this

Vsevolod [243]

Answer:

Explanation:

6 0

3 years ago

Write the molecular formula for the following compound.

Sholpan [36]

Answer : The molecular formula of the compound is, $C_8H_9O_3N$

Explanation :

Molecular formula : It is the representation of substance by the symbols and it denotes the number of atoms of each element present in the compound.

Now count the number of carbon, hydrogen, nitrogen and oxygen atoms present in the given compound.

As we see that in the given compound, there are 8 atoms of carbon element, 3 atoms of oxygen element, 1 atom of nitrogen element, 9 atoms of hydrogen element.

Thus, the molecular formula of the compound will be $C_8H_9O_3N$

7 0

3 years ago