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babunello [35]
3 years ago
8

Which of the following lists of elements are in the same PERIOD? *

Chemistry
1 answer:
yuradex [85]3 years ago
5 0
Ca, V, Cu, Kr
Remember a period is a the horizontal rows in the Periodic Table.

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How many grams of solid barium sulfate form when 25.0 ml of 0.160 m barium chloride reacts with 68.0 ml of 0.055 m sodium sulfat
GalinKa [24]
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Given the balanced equation representing a reaction at 101.3 kPa and 298 K:
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3 years ago
What electrostatic attraction between____ forms an ionic bond?
lina2011 [118]
Between atoms (one metall and one non metall) form an ionic bond(NaCl)
4 0
3 years ago
Be sure to answer all parts. Write the equations representing the following processes. In each case, be sure to indicate the phy
barxatty [35]

Answer:

1.S^{-}(g)+e^{-} \rightarrow S^{2-}(g)

2.Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}

3.The electron affinity of  Mg^{2+} is zero.

4.O^{2-}(g) \rightarrow O^{-}(g)+e^{-}

Explanation:

1.

<u>Electron affinity:</u>

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of S^{-} is as follows.

S^{-}(g)+e^{-} \rightarrow S^{2-}(g)

2.

<u>Ionization energy</u>:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}

3.

The electronic configuration of Mg: 1s^{2}2s^{2}2p^{6}3s^{2}

By the removal of two electrons from a magnesium element we get Mg^{2+} ion.

Mg^{2+} has inert gas configuration i.e,1s^{2}2s^{2}2p^{6}

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of  Mg^{2+} is zero.

4.

The ionization energy of O^{2-} is follows.

O^{2-}(g) \rightarrow O^{-}(g)+e^{-}

3 0
3 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
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