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3 years ago

The energy in eV for light with a wavelength of 6250 angstroms is _. Note - there are 1.6 x 10-12 erg in 1 eV.

Chemistry

1 answer:

vivado [14]3 years ago

5 0

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

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

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem we have:

$h=6.63\cdot 10^{-34} Js$

$c=3.0\cdot 10^8 m/s$

$\lambda=6250 A = 6250\cdot 10^{-10} m$ is the wavelength of the photon

Therefore, the energy in Joules is

$E=\frac{(6.63\cdot 10^{-34})(3.0\cdot 10^8)}{6250\cdot 10^{-10}}=3.2\cdot 10^{-19}J$

We want to convert this energy into electronvolts: we know that the conversion factor is

$1 eV = 1.6\cdot 10^{-19}J$

Therefore,

$E=\frac{3.2\cdot 10^{-19}}{1.6\cdot 10^{-19}}=2 eV$

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Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

How many milliliters of 0.1 m hcl is required to react with 0.15g sodium carbonate (na2co3?

docker41 [41]

1) Write the balanced chemical equation

2HCl + Na2 CO3 ----------> 2NaCl + H2CO3

2) Write the molar ratios:

2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3

3) Convert 0.15g of sodium carbonate to number of moles

3a) Calculate the molar mass of Na2CO3

Na: 2 * 23 g/mol = 46 g/mol

C: 12 g/mol =

O: 3 * 16 g/mol = 48 g/mol

molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol

3b.- Calculate the number of moles of Na2CO3

# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3

4) Calculate the number of moles of HCl from the molar proportion:

[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl

5) Calculate the volume of HCl from the definition of Molarity

Molarity, M = # moles / volume in liters

=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters

0.028 liters * 1000 ml / liter = 28 ml.

Answer: 28 mililiters of 0.1 M HCl.

7 0

3 years ago

A chemist prepares a solution of barium chloride by measuring out of barium chloride into a volumetric flask and filling the fla

mezya [45]

The given question is incomplete. The complete question is:

A chemist prepares a solution of barium chloride by measuring out 110 g of barium chloride into a 440 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mole per liter of the chemist's barium chloride solution. Round your answer to 3 significant digits.

Answer: Concentration of the chemist's barium chloride solution is 1.20 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times }{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

moles of $BaCl_2$ (solute) = $\frac{\text {given mass}}{\text {Molar Mass}}=\frac{110g}{208g/mol}=0.529mol$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.529\times 1000}{440ml}=1.20mole/L$

Therefore, the molarity of solution is 1.20 mol/L

8 0

3 years ago

What is the correct value for Avogadro's number?

nydimaria [60]

Answer: Option A. 6.022 x 10^23

Explanation:

8 0

3 years ago

A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

1 mole of aniline reacts per mole of HCl

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

7 0

3 years ago