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2 years ago

Which group on the periodic table contains chemical elements that are shiny and conduct electricity?

Chemistry

1 answer:

olga_2 [115]2 years ago

4 0

D. Metals
Elements that are shiny, solid, conducts electricity, has atomic mass of 108 is metal.

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Greenhouse gases in the atmosphere contribute to

lesantik [10]

Answer:

Explanation:

While solar power does produce carbon dioxide emissions in the manufacturing processes required to produce panels and batteries, it does not produce CO2 while converting solar energy to electrical energy.

6 0

2 years ago

The half life for the decay of radium is 1620 years what is the rate constant

Lilit [14]

Decay is a type of degradation reaction and thus is considered a first order reaction. thus the formula goes like this.

rate constant= 0.693/half life

so here...

rate constant= 0.693/1620 year^-1

3 0

3 years ago

A compound is found to contain 50. 05% sulfur and 49. 95% oxygen by mass. What is the empirical formula for this compound? SO S2

jekas [21]

The empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

The empirical formula has been the whole unit ratio of the elements in the formula unit.

<h3>Computation for the Empirical formula</h3>

The given mass of Sulfur has been, 50.05 g

The given mass of oxygen has been 49.95 g.

The moles of elements in the sample has been given by:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

Moles of Sulfur:

$\rm Moles\;S=\dfrac{50.05}{32}\\
 Moles\;S=1.56\;mol$

The moles of sulfur in the unit has been 1.56 mol.

Moles of Oxygen:

$\rm Moles\;O=\dfrac{49.95}{16} \\
Moles\;O=3.12\;mol$

The moles of oxygen in the unit has been 3.12 mol.

The empirical formula unit has been given as:

$\rm S_{1.56}O_{3.12}=SO_2$

Thus, the empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

Learn more about empirical formula, here:

brainly.com/question/11588623

8 0

2 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

How much heat (in kj) is released when 3.600 mol naoh(s) is dissolved in water? (the molar heat of solution of naoh is â445.1 kj

Fantom [35]

<span>-1602 kj

Sauce:me

Your welcome
</span>

8 0

3 years ago

Read 2 more answers