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3 years ago

A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

1 answer:

6 0

hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

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Which of the following displays the correct change in enthalpy and best describes the reaction below? 2CsCl(aq) + Na2SO4(aq) 2Na

Bond [772]

We subtract the enthalpies of the reactants from that of the products:
$2\Delta
 H(NaCl)+\Delta H(Cs_2 SO_4)-2\Delta H(CsCl)-\Delta H(Na_2 SO_4) \\ 
=2(-411)+(-1400) -2(-415)-(-1380) \\ = -12 kJ$
Since this is < 0, this is an exothermic reaction.

3 0

3 years ago

Which of the following is the correct set-up for the problem: How many grams of water will be produced from 3.2 moles of oxygen

suter [353]

Option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.

<h3>Reaction between oxygen and ethene</h3>

Ethene (C2H4) burns in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) along with the evolution of heat and light.

C₂H₄ + 3O₂ ----- > 2CO₂ + 2H₂O

from the equation above;

3 moles of O₂ ---------> 2(18 g) of water

3.5 moles of O₂ ----------> x

$x = 3.2 \times [\frac{2 \ moles \ H_2O}{3 \ moles \ O_2} ] \times[ \frac{18.02 \ g \ H_2O}{1 \ mole \ H_2O} ]$

Thus, option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.

Learn more about reaction of ethene here: brainly.com/question/4282233

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6 0

2 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

In 1887, Albert A. Michelson and Edward W. Morley conducted an experiment to measure the speed at which the Earth moves through

vichka [17]

C.The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, so the theory had to be rejected.

7 0

3 years ago

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a propane torch is lit inside a hot air balloon during preflight preparations to inflate the balloon. which condition of the gas

earnstyle [38]

Answer:

The pressure remains constant

Explanation:

this is an example in charles law where as the temperature increases so does the volume.

5 0

3 years ago

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