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3 years ago

A 500 mL gas sample is collected over water at a pressure of 740mmHg and 25 degrees C. What is the volume of the dry gas at STP?

Chemistry

1 answer:

Zina [86]3 years ago

8 0

Answer:

The volume of the dry gas at STP is = 0.4314 L

Explanation:

We are given:

Total vapor pressure = 740 mmHg

Also, considering Vapor pressure of water = 23.78 mmHg

Vapor pressure of gas = Total vapor pressure - Vapor pressure of water = (740 - 23.78) mmHg = 716.22 mmHg

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 716.22 mmHg

V = Volume of the gas = 500 mL = 0.5 L ( 1 ml = 0.001 L )

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of gas = ?

Putting values in above equation, we get:

$716.22mmHg\times 0.5L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{716.22\times 0.5}{62.3637\times 298}=0.01926mol$

At STP, 1 mole of gas yields a volume of 22.4 L

So,

0.01926 mole of gas yields volume of $0.01926\times 22.4\ L$

<u>Hence, the volume of the dry gas at STP is = 0.4314 L</u>

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determine the longest wavelength of light required to remove an electron from a sample of potassium metal, if the binding energy

Dafna11 [192]

Answer:

The wavelength of light is 68 nm.

Explanation:

Given data:

Binding energy of electron = 176 × 10³ Kj/ mol or (1.76× 10⁶ j/mol)

Wavelength of light require to remove the electron = ?

Solution:

E = hc / λ

h = planck constant (6.63×10⁻³⁴ J/s)

c = speed of light = (3×10⁸ m/s)

The energy require per electron is

1.76 × 10⁶ j / 6.02× 10²³ = 2.92 × 10⁻¹⁸ J

Now we will put the values in formula,

E = hc / λ

λ = hc / E

λ = 6.63×10⁻³⁴ m².kg.s⁻¹ . 3×10⁸ m/s / 2.92 × 10⁻¹⁸ J

λ = 19.89×10⁻²⁶ m /2.92 × 10⁻¹⁸ (m².kg.s⁻² = J)

λ = 6.8 ×10⁻⁸ m

λ = 6.8 ×10⁻⁸ × 10⁹

λ = 68 nm

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4 years ago

What is the formula for barium nitrate ?

brilliants [131]

Answer:

Ba(NO3)2

Explanation:

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3 years ago

How much heat must be removed from 456 g of water at 25.0°c to change it into ice at?

N76 [4]

This question can be simply solved by using heat formula,
Q = mCΔT

Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)

when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
Q = 456 g x 4.186 J / g K x 25 K
= 47720.4 J
= 47.72 kJ

hence 47.72 kJ of heat energy should be removed.

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3 years ago

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3 years ago

Read 2 more answers

If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a

Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

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3 years ago