How many moles is 145 liters of hydrogen gas at STP?

What is the name of Pb(NO3)2? Explain how you determined the bond type and the steps you used to determine the naming convention

Anastaziya [24]

The name of Pb(NO3)2 is lead (II) nitrate and its bond type is ionic bond. Details about bonding can be found below.

<h3>What is ionic bonding?</h3>

Ionic bond is a type of chemical bond where two atoms or molecules are connected to each other by electrostatic attraction.

According to this question, the chemical formula of lead (II) nitrate is given as follows: Pb(NO3)2.

The two constituent ions in this compound are as follows: Pb2+ and NO3-. They are bonded by an ionic bond.

Therefore, the name of Pb(NO3)2 is lead (II) nitrate and its bond type is ionic bond.

Learn more about ionic bond at: brainly.com/question/11527546

#SPJ1

3 0

2 years ago

Which protien founds in hair?

Inessa05 [86]

Keratin is found in hair

4 0

3 years ago

Determine the oxidation numbers of all the elements in the unbalanced reactions. Then, balance each redox reaction in a basic so

givi [52]

Answer for 1:

- Oxidation number of Mn in MnO4^-1= +7

- Oxidation number of O in MnO4^-1= -2

- Oxidation number of C in C2O4^-2= +3

- Oxidation number of O in C2O4^-2= -2

- Oxidation number of Mn in MnO2= +4

- Oxidation number of O in MnO2= -2

- Oxidation number of C in CO2= +4

- Balanced redox equation in basic solution:

$3C_2O_4^{-2}+2MnO_4^{-1}+4H_2O\operatorname{\rightarrow}6CO_2\text{ +}2MnO_2\text{ + }8OH^-$

Explanation for 1:

1st) To determine the oxidation numbers, it is necessary that the total sum of the charges is equal to that of the molecule or ion in the equation.

• Oxidation numbers in MnO4-,:

We know that the oxidation number for oxygen is -2, then we multiply it by the subscript 4 to find the whole charge of oxygen in this molecule.

Oxygen oxidation number: -2 x 4 = -8

Since the total charge of the molecule is -1, by difference, we will know that the oxidation number of Mn will be +7:

$\begin{gathered} Mn^{+7}O_4^{-2} \\ +7+[(-2*4)]=-1 \end{gathered}$

To confirm that the oxidation number that we determined exists for that element, we can check the Periodic Table of Elements.

We proceed in the same way with all molecules.

• Oxidation numbers in C2O4-2,:

$\begin{gathered} C_2^{+3}O_4^{-2} \\ (+3*2)+[(-2)*4]= \\ +6-8=-2 \end{gathered}$

• Oxidation numbers in MnO2,:

$\begin{gathered} Mn^{+4}O_2^{-2} \\ +4+[(-2)*2]= \\ +4-4=0 \end{gathered}$

• Oxidation numbers in CO2,:

$\begin{gathered} C^{+4}O_2^{-2}_ \\ +4+[(-2)*2]= \\ +4-4=0 \end{gathered}$

2nd) Now that we know the oxidation number os each atom in the reaction, then we can find the element that is oxidized and the element that is reduced.

We can see that Mn goes from +7 to +4, the Mn atom is reduced. And, the carbon atom goes from +3 to +4 so it oxidizes.

3rd) It is necessary to write the oxidation reaction and the reduction reaction separately and balancing all elements except oxygen and hydrogen:

Oxidation:

$C_2O_4^{-2}\text{ }\rightarrow\text{ 2}CO_2\text{ + 2}e^-_$

Reduction:

$MnO_4^{-1}\text{ + 3}e^-\rightarrow MnO_2$

4th) Since the reaction occurs in a basic solution, we must add water (H2O) to balance the oxygen atoms and hydroxyl ion (OH-) to balance the hydrogen atoms. In this case, the reduction reaction is the only one that needs to be balanced with water and hydroxyl ion.

$MnO_4^{-1}\text{ + 3}e^-\text{ +2}H_2O\operatorname{\rightarrow}MnO_2\text{ + 4}OH^-$

5th) It is necessary to balance the electrons in each half-reaction. So, we multiply each half-reaction by the number of electrons in the other half-reaction:

Oxidation:

$\begin{gathered} (C_2O_4^{-2}\operatorname{\rightarrow}\text{2}CO_2\text{ + 2}e_^-)*3 \\ 3C_2O_4^{-2}\operatorname{\rightarrow}6CO_2\text{ + 6}e^- \end{gathered}$

Reduction:

$\begin{gathered} (MnO_4^{-1}\text{ + 3}e^-\text{ + 2}H_2O\operatorname{\rightarrow}\text{ MnO}_2\text{ }+4OH^-)*2 \\ 2MnO_4^{-1}\text{ + 6}e^-\text{ + 4}H_2O\operatorname{\rightarrow}\text{ 2MnO}_2\text{ }+8OH^- \end{gathered}$

6th) We need to cancel out everything that is repeated on opposite sides of the reactions, including the electrons.

Finally, we can write the balanced redox equation:

$3C_2O_4^{-2}+2MnO_4^{-1}+4H_2O\operatorname{\rightarrow}6CO_2\text{ +}2MnO_2\text{ + }8OH^-$

6 0

1 year ago

How many moles of hydrogen gas are needed to produce 20 grams of hydrochloric acid (HCl)?

Charra [1.4K]

Answer:

0.27 mole

Explanation:

Data given:

mass of hydrochloric acid (HCl) = 20 g

moles of hydrogen (H₂O) = ?

Solution:

First we look to the reaction in which hydrogen react with Chlorine and make hydrochloric acid (HCl)

Reaction:

H₂ + Cl₂ -----------> 2HCl

Now look at the reaction for mole ratio

H₂ + Cl₂ -----------> 2HCl

1 mole 2 mole

So it is 1:2 mole ratio of hydrogen to hydrochloric acid (HCl)

As we Know

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Now convert moles to gram

H₂ + Cl₂ -----------> 2HCl

1 mole (2 g/mol) 2 mole (36.5 g/mol)

2 g 73 g

So,

we come to know that 2 g of hydrogen gives 73 g of hydrochloric acid (HCl) then how many grams of hydrogen will be required to produce 20 grams of hydrochloric acid (HCl).

Apply unity formula

2 g of H₂ ≅ 73 g of HCl

X g of H₂ ≅ 20 g of HCl

Do cross multiplication

X of H₂ = 2 g x 20 g / 73 g

X of H₂ = 0.55 g

Now convert grams of H₂ into moles

No. of moles = mass in grams/molar mass

Put values in above formula

No. of moles = 0.55 g / 2 (g/mol)

No. of moles = 0.27 mol

0.27 mole of hydrogen needed to produce 20 gram of hydrochloric acid (HCl).

7 0

3 years ago

H2O(g) → H2 + O2(g) Predict whether the reaction above would have an increase or a decrease in entropy.

nadya68 [22]

Since the reaction has one mole of reactants versus two moles of products, the reaction would have a decrease in entropy. One mole of gas exerts less pressure than two moles of gas, and therefore one mole of gas has more entropy than two moles of gas.

Hope this helps

5 0

4 years ago

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