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Problem Page Question It takes to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbo

Marizza181 [45]

This is a incomplete question. The complete question is:

It takes 348 kJ/mol to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon. Round your answer to correct number of significant digits

Answer: 344 nm

Explanation:

$E=\frac{Nhc}{\lambda}$

E= energy = 348kJ= 348000 J (1kJ=1000J)

N = avogadro's number = $6.023\times 10^{23}$

h = Planck's constant = $6.626\times 10^{-34}Js
$

c = speed of light = $3\times 10^8ms^{-1}$

$348000=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{348000}$

$\lambda=3.44\times 10^{-7}m=344nm$ $1nm=10^{-9}m$

Thus the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon is 344 nm

5 0

3 years ago

When a solution forms, what interactions between particles are involved? Which are exothermic? Which are endothermic?

Blizzard [7]

Answer:

Explanation:

There are three types of interactions involved between the particles when solution are formed.

1 : Solute - solute interaction:

2 : Solute - solvent interaction:

3 : Solvent - solvent interaction:

1 : Solute - solute interaction:

It is the inter-molecular attraction between the solute particles.

2 : Solute - solvent interaction:

It involve the inter-molecular attraction between solvent and solute particles.

3 : Solvent - solvent interaction:

It involve the intermolecular attraction between solvent particles.

Solutions are formed if the intermolecular attraction between solute particles are similar to the attraction between solvent particles.

Exothermic process:

The process will exothermic when solute solvent bonds are formed with the release of energy and energy required to brake the solute-solute particles and solvent solvent particles are less.

Endothermic process:

The process will be endothermic when energy required to break the solute-solute particles and solvent solvent particles are higher than energy released when solute solvent bonds are formed .

6 0

3 years ago

At 350°c, keq = 1.67 × 10-2 for the reversible reaction 2hi (g) ⇌ h2 (g) + i2 (g). what is the concentration of hi at equilibriu

mariarad [96]

According to the reversible reaction equation:

2Hi(g) ↔ H2(g) + i2(g)

and when Keq is the concentration of the products / the concentration of the reactants.

Keq = [H2][i2]/[Hi]^2

when we have Keq = 1.67 x 10^-2

[H2] = 2.44 x 10^-3

[i2] = 7.18 x 10^-5

so, by substitution:

1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2

∴[Hi] = 0.0033 M

7 0

3 years ago

Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago

In today's experiment, Solutions A and B are prepared as follows.

hram777 [196]

Answer:

D. 0.160

Explanation:

The solution A is obtained adding 2.0mL of a solution of bromocresol green, 5.0mL of 1.60M HAc and 2.0mL of a solution of KCl. The solution is diluted to 50mL

That means the HAc is diluted from 5.0mL to 50.0mL, that is:

50.0mL / 5.0mL = 10 times.

And the final concentration of HAc must be:

1.60M / 10 times =

0.160M

Right answer is:

3 0

3 years ago