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Please answer the following on the picture<br>ASAP PLZZZ

maxonik [38]

30. Atomic Number is 17

31. Chlorine

32. 35.45

33. Nonmetal

34. 3 energy levels

35. 7 Valence electrons

3 0

4 years ago

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

Semmy [17]

The balanced half-reaction for the product that forms at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻ → I₂ + 2e- and 2H₂O → O₂ + 4H⁺ + 4e⁻

<h3>What is Balanced Chemical Equation ?</h3>

The equation during which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation is called balanced chemical equation.

Now write the equation for FeI₂

At cathode:

Fe⁺² + 2e⁻ → Fe(s) Eo = - 0.44 V

2H₂O + 2e⁻ → H₂ + 2OH⁻ Eo = - 0.827 V

It is easy to decrease Fe⁺² ions than the water, the product which is formed at cathode is Iron.

At anode:

2I⁻ → I₂ + 2e- Eo = - 0.54 V

2H₂O → O₂ + 4H⁺ + 4e⁻ Eo = -1.23 V

O₂ gas formed at anode.

Thus from the above conclusion we can say that The balanced half-reaction for the product at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻ → I₂ + 2e- and 2H₂O → O₂ + 4H⁺ + 4e⁻.

Learn more about the Balanced chemical equation here: brainly.com/question/26694427

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6 0

2 years ago

50cm3 of a mixture of methane and hydrogen were mixed with excess oxygen and exploded, the product after cooling to the original

marin [14]

Answer:the initial composition of the reactants is

40cm^3 of CH4

40cm^3of H2

100cm^3 of H2O

Explanation:

Balanced reaction is

CH4 +H2+5/2O2______

CO2 +3H2O

Excess KOH at room temperature absorbs CO2 whose volume is given by 40cm^3 i.e the volume by which the solution decreases

So using Gay lussac combining ratio which states that gases combine in volumes that are in simple ratio to each other if gases.

Since CO2 in the equation is 1 mole

Means 1mole represent 40cm^3

So CH4:H2:O2 are in ratio of 1:1:5/2=(40:40:100)cm^3 respectively.

8 0

3 years ago

C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w

Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃ ΔH = –183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂ ΔH = 183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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7 0

2 years ago

Carbon monoxide (CO) is a poisonous gas because it binds very strongly to the oxygen carrier hemoglobin in blood. A concentratio

velikii [3]

<u>Answer:</u> The amount of CO that is occupied in the room is $1.98\times 10^3L$

<u>Explanation:</u>

We are given:

Concentration of CO = $8.00\times 10^2ppm=800pm$ by volume

This means that $800\mu L\text{ or }800\times 10^{-6}$ of CO is present in 1 L of blood

To calculate the volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of cuboid

l = length of cuboid = 10.99 m

b = breadth of cuboid = 18.97 m

h = height of cuboid = 11.89 m

$V=10.99\times 18.97\times 11.89=2478.83m^3$

Converting this into liters, by using conversion factor:

$1m^3=1000L$

So, $2478.83m^3=2.479\times 10^6L$

Applying unitary method:

In 1 L of blood, the amount of CO present is $800\times 10^{-6}$

So, in $2.479\times 10^6L$ of blood, the amount of CO present will be = $\frac{800\times 10^{-6}}{1}\times 2.479\times 10^{6}=1983.2L=1.98\times 10^3L$

Hence, the amount of CO that is occupied in the room is $1.98\times 10^3L$

8 0

3 years ago