Which of the following elements could be transmutated into silver by bombardment with a positron?

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

6 0

3 years ago

Match the characteristics and elements to their group.

Luba_88 [7]

Answer:

Kr is a Noble Gas. Na is an alkali metal. F is halogen.

Group 17 is halogens. Inert is Noble Gases. Odourless and colourless is Noble Gases. Alkali metals do not occur freely in nature. Alkali metals are malleable

Explanation:

8 0

3 years ago

Abbrevation of mass in the si unit

Leno4ka [110]

The SI unit for mass is Kilogram (kg).

3 0

4 years ago

Sodium hydroxide (NaOH) and hydrochloric acid (HCl) combine to make table salt (NaCl) and water (H2O). Which equation is balance

Art [367]

The equation that is balanced among the given options is the first equation NaOH + HCl → NaCl + H2O

From the question,

We are to determine which of the options gives a balanced equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) to make table salt (NaCl) and water (H₂O)

To do this, we will check which of the given equations is balanced, by checking the equations one after the other.

First, we will define what is meant by balanced equation.

Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products

For NaOH + HCl → NaCl + H2O

On the reactants side

Na = 1

O = 1

H = 2

Cl = 1

On the products side

Na = 1

O = 1

H = 2

Cl = 1

Since equal number of atoms of each element are present at both the reactants and products sides, then the equation is balanced

For 2NaOH + 2HCl → 2NaCl + H2O

On the reactants side

Na = 2

O = 2

H = 4

Cl = 1

On the products side

Na = 2

O = 1

H = 2

Cl = 2

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

For 2NaOH + HCl → NaCl + H2O

On the reactants side

Na = 2

O = 2

H = 3

Cl = 1

On the products side

Na = 1

O = 1

H = 2

Cl = 1

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

For NaOH + 2HCl → NaCl + H2O

On the reactants side

Na = 1

O = 1

H = 2

Cl = 2

On the products side

Na = 1

O = 1

H = 2

Cl = 1

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

Hence, the equation that is balanced is the first equation NaOH + HCl → NaCl + H2O

Learn more here: brainly.com/question/17015942

6 0

2 years ago

Read 2 more answers

Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH =

Mashcka [7]

<u>Answer:</u> The amount of heat released is -7.203 kJ

<u>Explanation:</u>

The given chemical equation follows:

$2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ$

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be $\frac{-196}{2}\times 1=-98kJ$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol$

To calculate the heat of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released

n = number of moles = 0.0735 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

$-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ$

Hence, the amount of heat released is -7.203 kJ

8 0

4 years ago