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3 years ago

Which atom in the ground state has five electrons in its outee level and ten electrons in its kernel (1) C (2) Cl (3) Si (4) P

1 answer:

4 0

If you use its electronic structure the last number should be a five so something like (2,5), (2,8,5) or (2,8,8,5) etc.

The elements therefore with 5 as it's final number and 5 on the outer orbit are N, P, As, Sb, Bi, Uup. Therefore with the answers given to you the answer is most likely 4) Phosphorus. <span />

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The abundance of coyotes has led to the killing of several cows

Vilka [71]

I think variation.... have a great day

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3 years ago

Read 2 more answers

4. How many milligrams are in 5.25 x 10-13 kg?<br><br> the “-13” is an exponent

rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

Given mass = 5.25 x 10⁻¹³kg

The kilo- is a prefix that denotes 10³

therefore;

1000g = 1kilogram

the milli- is a prefix that denotes 10⁻⁻³

1000mg = 1g

Now that we know this, we can convert:

5.25 x 10⁻¹³kg x $\frac{1000g}{1kg} x \frac{1000mg}{1g}$ = 5. 25 x 10⁻¹³ x 10⁶mg

= 5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

8 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Calculate the density of a solid in g/cm3 if it weighs 38.3 kg and has a volume of 0.00463 m3.

RideAnS [48]

Answer: D=8.27 g/cm³

Explanation:

Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.

$D=\frac{38.3kg}{0.00463m^3} *\frac{(1m)^3}{(100cm)^3} *\frac{1000g}{1kg}$

With this equation, the m³ and kg cancel out, and we are left with g/cm³.

D=8.27 g/cm³

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3 years ago

HELP FAST!!!

AfilCa [17]

Answer:

the fourth one I think....m

Explanation:

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3 years ago