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Greeley [361]
2 years ago
14

Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F

alse" to indicate whether or not the stated reaction (or "no reaction") correctly corresponds to the expected observation in each case. Lead(II) nitrate and sodium chloride; No reaction occurs. Sodium bromide and hydrochloric acid; No reaction occurs. Nickel(II) chloride and lead(II) nitrate; Pb2+(aq) + 2Cl-(aq) --> PbCl2(s) Magnesium chloride and sodium hydroxide; No reaction occurs. Ammonium sulfate and barium nitrate; Ba2+(aq) + SO42-(aq) --> BaSO4(s)
Chemistry
1 answer:
nirvana33 [79]2 years ago
3 0

<u>Answer:</u>

<u>For 1:</u> The correct answer is False.

<u>For 2:</u> The correct answer is True.

<u>For 3:</u> The correct answer is True.

<u>For 4:</u> The correct answer is False.

<u>For 5:</u> The correct answer is True.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.  If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

  • <u>For 1:</u> Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is False.

  • <u>For 2:</u> Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)

Ionic form of the above equation follows:

Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

  • <u>For 3:</u> Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is True.

  • <u>For 4:</u> Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the correct answer is False.

  • <u>For 5:</u> Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)

Hence, the correct answer is True.

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8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
  • The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
  • 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
  • The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

5 0
2 years ago
How many total ions are present in 347g of cacl2?
vladimir1956 [14]

In 1 mole of CaCl_{2}, there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of Cl^{-1}.

CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-}

Molar mass of CaCl_{2} is 110.98 g/mol. Calculating number of moles from given mass as follows:

n=\frac{m}{M}=\frac{347 g}{110.98 g/mol}=3.12 mol

Thus, number of moles of ions will be 3\times 3.12 mol=9.38 mol.

Since, 1 mole of any substance has 6.023\times 10^{23} units of that substance where 6.023\times 10^{23}  is Avogadro's number.

Thus, 9.38 mol of ions will have 9.38\times 6.023\times 10^{23}=5.65\times 10^{24} number of ions.

Therefore, total number of ions in 347 g of CaCl_{2} is  5.65\times 10^{24}.


8 0
2 years ago
Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce
sergeinik [125]

Answer:

The empirical formula is = C_4H_8O

The formula of Valproic acid = C_8H_{16}O_2

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of H_2O = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

<u>Moles of H = 2 x 0.00922 = 0.01844 moles </u>

Molar mass of H atom = 1.008 g/mol

<u>Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g </u>

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of CO_2 = 0.403 g  /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

<u>Moles of C = 0.009157 moles </u>

Molar mass of C atom = 12.0107 g/mol

<u>Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g </u>

<u>Given that the Valproic acid only contains hydrogen, oxygen and carbon. </u>So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 0.165 g

<u>Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g  </u>

Molar mass of O = 15.999 g/mol

<u>Moles of O  = 0.036412  / 15.999  = 0.002276 moles</u>

<u></u>

<u>Taking the simplest ratio for H, O and C as: </u>

<u>0.01844 : 0.002276 : 0.009157</u>

<u> = 8 : 1 : 4</u>

The empirical formula is = C_4H_8O

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,  

Molecular mass = n × Empirical mass

144 = n × 72

<u>⇒ n = 2</u>

The formula of Valproic acid = C_8H_{16}O_2

7 0
2 years ago
Calculate the wavelength of the photon emitted when an electron makes a transition from n=6 to n=3. You can make use of the foll
Angelina_Jolie [31]

<u>Answer:</u> The wavelength of light is 1.094\times 10^{-6}m

<u>Explanation:</u>

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.097\times 10^7m^{-1}

n_f = Final energy level = 3

n_i = Initial energy level = 6

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{6^2} \right )\\\\\lambda =\frac{1}{914617m^{-1}}=1.094\times 10^{-6}m

Hence, the wavelength of light is 1.094\times 10^{-6}m

6 0
2 years ago
Aluminum chlorohydrate, al2(oh)5cl, is an active ingredient in some antiperspirants. what is the mass percent of aluminum in thi
Flauer [41]
The percet of Al will be 100 times the mass of Al in the formula divided by the molar mass of the compound.

This table shows all the calculations involved

Compound: Al2 (OH)5 Cl

element     # of        atomic      mass in the          %
                  atoms    mass        formula
                                g/mol

Al              2             27             2*27 = 54 g        (54 / 174.5)*100 = 30.9%

O              5            16              5*16 = 80 g

H              5             1               5*1 =     5 g

Cl             1             35.5          1*35.5 = 35.5 g
                                              ----------------------
                   molar mass                      174.5 g

Answer: 30.9%
7 0
2 years ago
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