Chemistry, 21.04.2020 19:26 youngchapo813p8d9u1

3. Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon). (6 pts)

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Answer from: rosehayden21

the theoretical value for the number of moles of $CO_{2(aq)}$ is 0.0173 moles

Explanation:

The balanced chemical equation for the reaction is represented by:

$H_3C_6H_5O_{7(aq)} + 3NaHCO_{3(aq)} ------3CO_{2(g)}+3H_2O+Na_3C_6H_5O_{7(aq)}$

From above equation; we would realize that 3 moles of $NaHCO_{3(aq)}$ reacts with $H_3C_6H_5O_{7(aq)}$ to produce 3 moles of $CO_{2(aq)}$

However ; the molar mass of $NaHCO_{3(aq)}$ = 84 g/mol

mass given for $NaHCO_{3(aq)}$ = 1.45 g

therefore , we can calculate the number of moles of $NaHCO_{3(aq)}$ by using the expression :

number of moles of $NaHCO_{3(aq)}$ = $\frac{mass \ given}{ molar \ mass}$

number of moles of $NaHCO_{3(aq)}$ = $\frac{1.45}{84}$

number of moles of $NaHCO_{3(aq)}$ = 0.0173 mole

Since the ratio of $NaHCO_{3(aq)}$ to $CO_{2(aq)}$ is 1:1; that implies that number of moles of $NaHCO_{3(aq)}$ is equal to number of moles of $CO_{2(aq)}$ produced.

number of moles of $CO_{2(aq)}$ = $\frac{mass \ given}{ molar \ mass}$

0.0173 = $\frac{mass \ given}{ 44 \ g/mol}$

mass of $CO_{2(aq)}$ = 0.0173 × 44

mass of $CO_{2(aq)}$ = 0.7612 g

Thus; the theoretical value for the number of moles of $CO_{2(aq)}$ is 0.0173 moles

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

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