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3 years ago

10

To break the oxygen-oxygen bonds in a single O2 molecule, 1 × 10–18 J of energy is required. Which of the following wavelengths

of light could be used to carry out this process?

1 answer:

Sonbull [250]3 years ago

8 0

E = hc/(lamda)

The lamda symbol is wavelength, which this site does not have. I can represent it with an "x" instead.

Plancks constant, h = 6.626×10^-32 J·s

Speed of light, c = 3.00×10^8 m/s

The energy must be greater than or equal to 1×10^-18 J

1×10^-18 J ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / x

x ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / (1×10^-18 J)

x ≤ 1.99×10^-7 m or 199 nm

The wavelength of light must be greater than or equal to 199 nm

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3 years ago

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You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. Aft

olya-2409 [2.1K]

Answer : The enthalpy of reaction $(\Delta H_{rxn})$ is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

$Mass=Density\times Volume$

Volume of solution = Volume of HCl + Volume of $Ba(OH)_2$

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

$Mass=1g/mL\times 93.1mL=93.1g$

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat released = ?

m = mass = 93.1 g

$C_p$ = specific heat capacity of water = $4.184J/g^oC$

$T_1$ = initial temperature = $25.0^oC$

$T_2$ = final temperature = $29.83^oC$

Now put all the given value in the above formula, we get:

$Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC$

$Q=1881.43J=1.88kJ$ (1 kJ = 1000 J)

Now we have to calculate the moles of $Ba(OH)_2$ and HCl.

$\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}$

$\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}$

$\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $Ba(OH)_2$ react with 2 mole of $HCl$

So, $9.71\times 10^{-3}$ moles of $Ba(OH)_2$ react with $9.71\times 10^{-3}\times 2=0.0194$ moles of $HCl$

From this we conclude that, $HCl$ is an excess reagent because the given moles are greater than the required moles and $Ba(OH)_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $Ba(OH)_2$ react to give 2 mole of $H_2O$

So, $9.71\times 10^{-3}$ moles of $Ba(OH)_2$ react with $9.71\times 10^{-3}\times 2=0.0194$ moles of $H_2O$

Now we have to calculate the change in enthalpy of the reaction.

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

where,

$\Delta H_{rxn}$ = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

Now put all the given values in above expression, we get:

$\Delta H_{rxn}=-\frac{1.88kJ}{0.0194mole}=-96.9kJ/mole$

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction $(\Delta H_{rxn})$ is, -96.9 kJ/mole

3 0

4 years ago

When an atom of the unstable isotope Na-24 decays, it becomes an atom of Mg-24 because the Na-24 atom spontaneously releases:(1)

Yanka [14]

When an atom of the unstable isotope Na-24 decays, it becomes an atom of Mg-24 because the Na-24 atom spontaneously releases (2) a<span> beta particle</span>.

4 0

3 years ago

When a 17.7 mL sample of a 0.368 M aqueous hypochlorous acid solution is titrated with a 0.301 M aqueous barium hydroxide soluti

nordsb [41]

Answer:

pH = 12.98

Explanation:

Step 1: Data given

Volume of aqueous hypochlorous acid solution = 17.7 mL = 0.0177 L

Molarity of aqueous hypochlorous acid solution = 0.368 M

Molarity of aqueous barium hydroxide solution = 0.301 M

Volume of aqueous barium hydroxide solution = 16.2 mL = 0.0162 L

Step 2: The balanced equation

2HCl + Ba(OH)2 → BaCl2 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles HCl = 0368 M * 0.0177 L

Moles HCl = 0.0065136 moles

Moles Ba(OH)2 = 0.301 M * 0.0162 L

Moles Ba(OH)2 = 0.0048762 moles

Step 4: Calculate the limiting reactant

For 2 moles HCl we need 1 mol Ba(OH)2 to produce 1 mol BaCl2 and 2 moles H2O

HCl is the limiting reactant. It will completely be consumed 0.0065136 moles. Ba(OH)2 is in excess. There will react 0.0065136/2 = 0.0032568‬ moles. There will remain 0.0048762 moles - 0.0032568‬ = 0.0016194 moles

Step 5: Calculate molarity Ba(OH)2

Molarity Ba(OH)2 = moles / volume

Molarity Ba(OH)2 = 0.0016194 moles / 0.0339 L

Molarity Ba(OH)2 = 0.04777 M

Step 6: Calculate [OH-]

Ba(OH)2 → Ba^2+ + 2OH-

For Ba(OH)2 we have 2* [OH-]

[OH-] = 2*0.04777 = 0.09554 M

Step 7: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.09554)

pOH = 1.02

Step 8: Calculate pH

pH = 14 - 1.02

pH = 12.98

8 0

3 years ago

Which term names a metamorphic change in rocks over a wide area from tectonic forces in Earth's crust and mantle? A. local metam

Pavlova-9 [17]

D is the answer. A, B,C are absurd. Doing a little word right there.

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3 years ago

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