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Talja [164]
3 years ago
9

Rubidium has a mass number of 87 and an atomic number of 37. how many electrons does it have?

Chemistry
2 answers:
Elis [28]3 years ago
5 0
Rubidium has 37 elections. there are multiple other sources you could use too confirm this. :-)
Aleksandr [31]3 years ago
5 0
Rubidium's (Rb) Z = 37
Z ( atomic number ) means , No: of proton or no: of electrons.
.: Number of electrons = 37
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An element’s atomic number is the?
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You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35. How many atoms of lead are in the samp
aliya0001 [1]

<u>11.35 g</u><u> </u><u>atoms </u><u>of lead are in the sample.</u>

What does density mean?

  • The density of an area refers to the quantity of things—which may include people, animals, plants, or objects—there are in it.
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We can calculate the volume of the cube by cubing the side of the cube.

Volume of cube = ( 1.000 cm)³

                           = 1.000 cm³

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           =  ( 1.000 cm³) ( 11.35 \frac{g}{cm^{3} } )

         = 11.35 g

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3 0
1 year ago
In terms of energy what kind of system is a sealed glass terrarium on a windsill?
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6 0
3 years ago
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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
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