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2 years ago

Write the balanced chemical equation for the reaction between aqueous sodium carbonate and

Chemistry

1 answer:

ioda2 years ago

7 0

Answer:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

Explanation:

This question is asking to write and balance an equation between between aqueous sodium carbonate (Na2CO3) and aqueous nitric acid (HNO3). The equation is as follows:

HNO3 (aq) + Na2CO3 (aq) → NaNO3 (aq) + CO2 (g) + H2O (l)

However, this equation is not balanced as the number of atoms of each element must be the same on both sides of the equation. To balance the equation, one will make use of coefficients as follows:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

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An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

Which material has a crystalline structure at room temperature ( 20 degrees Celsius )

Sauron [17]

It's 2, glass. Water, nitrogen, and sucrose don;t have a crystalline structure.

8 0

2 years ago

Read 2 more answers

Which statement below matches the correct response with the proper reasoning when comparing the volatility of CH2Cl2 with CH2Br2

Nutka1998 [239]

Answer:

b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂

Explanation:

CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE

Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.

There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.

I hope it helps!

5 0

2 years ago

6. The graph below shows the heating curve for ethanol (from –200C to 150C). Calculate the amount of heat (kJ) required for each

Kazeer [188]

This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:

<h3>Heating curves:</h3>

In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.

Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

$Q_T=Q_1+Q_2+Q_3+Q_4+Q_5$

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:

$Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\
\\
Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\
\\
Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\
\\
Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\
\\
Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ$

Finally, we add them up to get the result:

$Q_T=0.208kJ+0.106kJ+0.470kJ+0.838kJ+0.136kJ\\
\\
Q_T=1.758kJ$

Learn more about heating curves: brainly.com/question/10481356

7 0

1 year ago

Typically if oxygen has a formal negative charge, it is participating in ____ covalent bond(s). Report your answer as a whole nu

ZanzabumX [31]

Oxygen is participating in non-polar covalent bond(s)

Oxygen forms 2 covalent bond.

This is because oxygen atoms have 6 valence electrons. This means that it has 2 lone pairs and 2 unpaired electrons that are shared in order to achieve octet configuration.

In this chemistry, the 2 lone pairs on the oxygen are not shared with any other atoms. Instead; they are assigned to the oxygen atom. The formal charge on the oxygen atom is zero. Oxygen's atomic number is 8 and is equal to the sum of the number of its valence and inner shell electrons.

<h3>What is non-polar covalent bonding?</h3>

Nonpolar covalent bonding is a kind of covalent bond in which the bonding electrons are shared equally between the two atoms.

Learn more about non-polar covalent bonding:

brainly.com/question/26056508

8 0

2 years ago