The number of __________ in an atom’s nucleus is equal to its __________

Electrochemical processes are written with the oxidation half-reaction on the left and the reduction reaction on the right; wher

CaHeK987 [17]

Answer:

B) FALSE.

Explanation:

First off, its important to understand the following concepts;

A half reaction is either the oxidation or reduction reaction component of a redox (Oxidation - Reduction) reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.

Often, the concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).

Half equations simply much just breaks the reaction into oxidation and reduction steps (irrespective of the order, whether left or right).

An example of half equation is given below;

Mg → Mg2+ + 2e− (Oxidation)

Cu2+ + 2e− → Cu (Reduction)

The answer is false because; oxidation-reduction reactions are NOT represented by equations that group the reaction participants into reactants and products. Rather they are

8 0

3 years ago

The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this

Semmy [17]

Answer:

The molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

, $=\frac{0.07}{100}\times 21.5}\\=0.0151g$ of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

$\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M$

The chemical equation for the ionization of copper (II) sulfate follows:

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = $4.73\times 10^{-5}M$

Hence, the molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

4 0

3 years ago

How can you write Mg(OH)2 in subscripts

Liula [17]

Mg(OH)2 is the correct chemical formula because it takes two Hydroxide ions to bond with Magnesium normally it would be written with a subscript number 2 rather than the parenthesis and number 2

7 0

3 years ago

In an experiment designed to determine the concentration of Cu 2 ions in an unknown solution, you need to prepare 100 mL of 0.10

Alexxandr [17]

Answer:

1.6 grams

Explanation:

We need to prepare 100 mL (0.100 L) of a 0.10 M CuSO₄ solution. The required moles of CuSO₄ are:

0.100 L × 0.10 mol/L = 0.010 mol

The molar mass of CuSO₄ is 159.61 g/mol. The mass corresponding to 0.010 moles is:

0.010 mol × (159.61 g/mol) = 1.6 g

We should use 1.6 grams of CuSO₄.

8 0

3 years ago

How many liters of solvent are needed to make a 6.5M solution with 34 mol of solute?

Dominik [7]

Answer:

5.231 L.

Explanation:

Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(Volume of the solution (L))</em>

M = 6.5 M.

no. of moles of solute = 34.0 mol,

Volume of the solution = ??? L.

∴ (6.5 M) = (34.0 mol)/(Volume of the solution (L))

∴ (Volume of the solution (L) = (34.0 mol)/(6.5 M) = 5.231 L.

6 0

3 years ago

The number of __________ in an atom’s nucleus is equal to its ___________ number?

The number of in an atom’s nucleus is equal to its _ number?