What ALWAYS occurs simultaneously with reduction?

Science.... <br> homework!

Leokris [45]

The answer to this question is 2 and 3

5 0

4 years ago

In which substance does phosphorus have a +3 oxidation state?

Mkey [24]

Examples: phosphorus trichloride
( PCl3)
Phosphorus trioxide
( P2O3)

4 0

3 years ago

For a particular reaction, ΔH⁰ = − 46.7 kJ/mol and ΔS⁰ = − 123.8 J / (mol ⋅ K). Assuming these values change very little with te

PilotLPTM [1.2K]

Answer:

T = 377.2 K, Less than

Explanation:

The thermodynamic quantity used in predicting whether a reaction is spontaneous or not is the gibbs free energy.

It's relationship with ΔH⁰ and ΔS⁰ is given as;

ΔG° = ΔH° - TΔS°

Basically, a negative value of ΔG° means the reaction is spontaeneous.

To obtain the calculated vale of T,

ΔS° = ΔH°/T

T = ΔH° / ΔS°

T = 377.2 K

Let's calculate the value of ΔG° at that temperature.

ΔG° = ΔH° - TΔS°

ΔG° = − 46700 - 377.2(− 123.8)

ΔG° = 0 (approximately, values are due to the rounding off)

At ΔG° = 0 the reaction is at equilibrium.

To find if the reaction is spontaneous at lower or hugher temperature than the calculated temperature, we would be substituting the value of T with a smaller (random) value and also a larger (random) value.

Larger T (390K)

ΔG° = ΔH° - TΔS°

ΔG° = − 46700 - 390(− 123.8)

ΔG° = - 46700 + 48,282

ΔG° = 1582 J/mol

Smaller T (350K)

ΔG° = ΔH° - TΔS°

ΔG° = − 46700 - 350(− 123.8)

ΔG° = - 46700 + 43330

ΔG° = -3370J/mol

This means the temperature would be lesser than the calculated value for it to be spontaneus.

8 0

3 years ago

A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the io- dide i

Sergeeva-Olga [200]

Answer:

$m_{I^-}=1.18gI^-$

Explanation:

Hello,

In this case, the reaction is given as:

$I^-+AgNO_3\rightarrow AgI+NO_3^-$

Thus, starting by the yielded grams of silver iodide, we obtain:

$n_{I^-}=2.185gAgI*\frac{1molAgI}{234.77gAgI}*\frac{1molI}{1molAgI}=9.31x10^{-3}molI^-$

Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:

$m_{I^-}=9.31x10^{-3}molI^-\frac{127gI^-}{1molI^-} =1.18gI^-$

And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.

Best regards.

7 0

4 years ago

1. Calculate the molarity when 403 g MgSO4 is dissolved to make a 5.25 L solution. Round to two significant digits.

lisov135 [29]

Answer:

1. Molarity of MgSO₄ = 0.6 M

2. Molarity of AgNO₃ = 0.06 M

3. volume of acetic acid = 250 mL

4. Molarity of NaCl= 2.3 M

5. Mass of C₁₂H₂₂O₁₁ = 4.1 g

6. Mass of C₁₀H₈ (naphthalene) = 77 g

7. mass of ethanol = 11 g

8. Mass of DDT = 0.011 mg

Explanation:

Ans 1.

Data given

mass of MgSO₄ = 403

Volume of solution = 5.25 L

Solution:

First we will find mole of solute

no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of MgSO₄ = 24 + 32 + 4(16)

Molar mass of MgSO₄ = 120 g/mol

Put values in equation 1

no. of moles = 403 g / 120 g/mol

no. of moles = 3.36 mol

Now to find molarity

Formula used

Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

M = 3.36 mol / 5.25 L

M = 0.64

So, round the figure to two significant digits.

Molarity of MgSO₄ = 0.64 M

_____________________

Ans 2.

Data given

mass of AgNO₃ = 1.24 g

Volume of solution = 125 mL

convert mL to L

1000 mL = 1 L

125 mL = 125/1000 = 0.125

Solution:

First we will find mole of solute

no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of AgNO₃ = 108 + 14 + 3(16)

Molar mass of AgNO₃ = 170 g/mol

Put values in equation 1

no. of moles = 1.24 g / 170 g/mol

no. of moles = 0.0073 mol

Now to find molarity

Formula used

Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

M = 0.0073 mol / 0.125 L

M = 0.06

So, round the figure to two significant digits.

Molarity of AgNO₃ = 0.06 M

______________________

Ans 3.

Data Given:

volume of acetic acid = 500 mL

% solution of acetic acid (M)= 50%

volume of acetic acid needed = ?

Solution:

formula used

percent of solution = volume of solute/ volume of solution x 100

Put Values in above formula

50 % = volume of solute / 500 mL x 100

Rearrange the above equation

volume of solute = 50 x 500 mL /100

volume of solute = 250 mL

So, volume of acetic acid = 250 mL

______________________

Ans 4

Data Given:

Molarity of NaCl (M1) = 6 M

Volume of NaCl (V1) = 750 mL

convert mL to L

1000 ml = 1 L

750 ml = 750/1000 = 0.75 L

Volume of dilute solution (V2)= 2 L

Molarity of dilute solution (M2) = ?

Solution:

Dilution Formula will be used

M1V1 = M2V2 . . . . . . (1)

Put values in equation 1

6 M x 0.75 L = M2 x 2 L

Rearrange the above equation

M2 = 6 M x 0.75 L / 2 L

M2 = 2.25 M

So, round the figure to two significant digits.

Molarity of NaCl= 2.3 M

_________________________

Ans 5.

Data Given:

Molarity of C₁₂H₂₂O₁₁ = 0.16 M

Mass of C₁₂H₂₂O₁₁ (m) = ?

Volume of dilute solution = 75 mL

convert mL to L

1000 ml = 1 L

75 ml = 75/1000 = 0.075 L

Solution:

As we know

Molarity = no.of moles/ liter of solution . . . . . . .(1)

we also know that

no.of moles = mass in grams / molar mass . . . . . (2)

Combine both equation 1 and 2

Molarity = (mass in grams / molar mass) / liter of solution . . . . (3)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

Put values in equation in equation 3

0.16 M = (mass in grams / 342 g/mol) / 0.075 L

Rearrange the above equation

mass in grams = 0.16 mol/L x 0.075 L x 342 g/mol

mass in grams = 4.104 g

So, round the figure to two significant digits.

Mass of C₁₂H₂₂O₁₁ = 4.1 g

____________________

The remaing portion is in attachment

8 0

4 years ago