When the elements in Group 1 are considered in order from top to bottom, each successive element at standard pressure has

Write the balanced molecular and net ionic equation for the reaction that occurs when the contents of the two beakers are added

hram777 [196]

1) hydrochloric acid with nickel:

Balanced molecular equation: Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g).

Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

Nickel is oxidized from oxidation number 0 (Ni) to oxidation number +2, hydrogen is reduced from oxidation number +1 to oxidation number 0 (H₂).

2) sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂.

Iron is oxidized from oxidation number 0 (Fe) to oxidation number +2, hydrogen is reduced from oxidation number +1 to oxidation number 0 (H₂).

3) hydrobromic acid with magnesium :

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

Magnesium is oxidized from oxidation number 0 (Mg) to oxidation number +2, hydrogen is reduced from oxidation number +1 to oxidation number 0 (H₂).

4) acetic acid with zinc:

Balanced molecular equation: Zn(s) + 2CH₃COOH(aq) → (CH₃COO)₂Zn(aq) + H₂(g).

Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Zinc is oxidized from oxidation number 0 (Zn) to oxidation number +2 (Zn²⁺), hydrogen is reduced from oxidation number +1 to oxidation number 0 (H₂).

7 0

2 years ago

In the electrochemical cell using the redox reaction below, the anode half reaction is ________. Sn4+ (aq) + Fe (s) → Sn2+ (aq)

kenny6666 [7]

Answer:

The anode half reaction is : $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}$

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}$

Reduction (cathode): $Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)$

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overall: $Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)$

So the anode half reaction is : $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}$

8 0

2 years ago

The balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver chloride. NaCl + Ag

Verdich [7]

<span>NaCl First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound Sodium = 22.989769 Chlorine = 35.453 Silver = 107.8682 Nitrogen = 14.0067 Oxygen = 15.999 Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results For NaCl 22.989769 + 35.453 = 58.44277 For AgNO3 107.8682 + 14.0067 + 3 * 15.999 = 169.8719 Now calculate how many moles of each substance by dividing the total mass by the molar mass For NaCl 4.00 g / 58.44277 g/mol = 0.068443 mol For AgNO3 10.00 g / 169.8719 g/mol = 0.058868 Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>

8 0

2 years ago

A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.7 nm.93.7 nm. What is the final state of the hy

sammy [17]

Answer:

5

Explanation:

Given that the formula is;

1/λ= R(1/nf^2 - 1/ni^2)

λ = 93.7 nm or 93.7 * 10^-9 m

R= 1.097 * 10^7 m-1

nf = ?

ni = 1

From;

ΔE = hc/λ

ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9

ΔE = 21 * 10^-19 J

ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)

-0.963 = (1/nf^2 - 1)

-0.963 + 1 = 1/nf^2

0.037 = 1/nf^2

nf^2 = (0.037)^-1

nf^2 = 27

nf = 5

7 0

2 years ago

There are ________ hydrogen atoms in 25 molecules of c4h4s2.

coldgirl [10]

Answer: 100.

Explanation:

1) The subscripts to the right of each element (symbol) in the chemical formula tells the number of atoms of that element present in one unit formula.

2) The unit formula of C₄H₄S₂ is equal to 1 molecule.

3) Therefore, there are 4 carbon atoms, 4 hydrogen atoms and 2 sulfur atoms in each molecule of C₄H₄S₂.

4) Then, you just have to multiply the corresponding subscript of the element times the number of molecules (25 in this case) to find the number of atoms of that kind.

5) These are the calculations for each element in the molecule C₄H₄S₂.

i) C: 4 × 25 = 100

ii) H: 4 × 25 = 100

iii) S: 2 × 25 = 50.

6) The question is about H only, so the answer is that there are 100 hydrogen atoms in 25 molecules of C₄H₄S₂.

5 0

2 years ago

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