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3 years ago

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Young's experiment is performed with light from excited helium atoms (λ = 496 nm). fringes are measured carefully on a screen 1.

25 m away from the double slit, and the center of the twentieth fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. what is the separation of the two slits?

1 answer:

Lunna [17]3 years ago

6 0

Ill answer in a second hold on plz.

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A 1.00 g sample of a hydrogen peroxide (H2O2) solution is placed in an Erlenmeyer flask and diluted with 20 mL of 1 M aqueous su

mihalych1998 [28]

The percent by mass is0.613

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3 years ago

Consider two bulbs seperated by a valce. Both bulbs are amintained at the same temperature. Assume that when the valve between t

Juliette [100K]

Answer:

The pressure is $P_f = 0.93 \ atm$

Explanation:

From the question we are told that

The volume of Ne is $V_N = 2.50 \ L$

The volume of CO is $V_C = 2.00 \ L$

The pressure of $Ne$ is $P_N = 1.09 \ atm$

The pressure of CO is $P_C = 0.773 \ atm$

The number of moles of Ne present is evaluated using the ideal gas equation as

$n_N = \frac{P_N * V_N}{R T}$

=> $n_N = \frac{1.09 * 2.50 }{R T} = \frac{2.725}{RT}$

The number of moles of CO present is evaluated using the ideal gas equation as

$n_N = \frac{P_C * V_C}{R T}$

=> $n_N = \frac{0.73 * 2.00 }{R T} = \frac{1.46}{RT}$

The total number of moles of gas present is evaluated as

$n_T = n_N + n_C$

$n_T = \frac{2.725}{RT} + \frac{1.46}{RT}$

$n_T = \frac{4.185}{RT}$

The total volume of gas present when valve is opened is mathematically represented as

$V_T = V_N + V_C$

=> $V_T = 2.50 + 2.00 = 4.50 \ L$

So

From the ideal gas equation the final pressure inside the system is mathematically represented as

$P_f = \frac{n_T * RT }{ V_T}$

=> $P_f = \frac{[\frac{4.185}{RT} ] * RT }{ 4.50}$

=> $P_f = 0.93 \ atm$

3 0

3 years ago

If an object has a density of 0.55 g/mL, what is its density in cg/L?

otez555 [7]

if 1 g is equal to 100 cg

then 0.55 g are equal to X cg

X = (0.55 × 100 ) / 1 = 55 cg

The density of the object is 55 cg/L.

7 0

3 years ago

Read 2 more answers

The solubility of nitrogen gas at 25 degrees C and 1 atm is 6.8 x 10^(-4) mol/L. If the partial pressure of nitrogen gas in air

Zolol [24]

Answer:

5.2 x 10⁻⁴ M.

Explanation:

The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:

<em>P = kC</em>

where P is the partial pressure of the gaseous solute above the solution.

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas.

At two different pressures, there is two different concentrations of dissolved gases and is expressed in a relation as:

<em>P₁C₂ = P₂C₁,</em>

P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.

P₂ = 0.76 atm, C₂ = ??? mol/L.

<em>∴ C₂ = (P₂C₁)/P₁ =</em> (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = <em>5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.</em>

5 0

3 years ago

An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0

3 years ago