I) __________are rigid due to strong (ii)___________ of attraction between the particles

You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th

ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially, only 4 moles of oxygen gas were present in the flask.

$p_{O_2}=Tp_1\times X_{O_2}$ ( $X_{O_2}=\frac{4}{4}$ ) ( according to Dalton's law of partial pressure)

$p_{O_2}=Tp_1\times 1=Tp_1$ ....(1)

$Tp_1$ = Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

$X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}$

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

$p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

$Tp_2$ = Total pressure of the mixture.

from (1)

$Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

On rearranging, we get:

$Tp_2=2\times Tp_1$

The new total pressure will be twice of initial total pressure.

7 0

2 years ago

Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7

brilliants [131]

Answer: $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$

Explanation:

$HF\rightarrow H^+F^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.056 M and $\alpha$ = ?

$K_a=1.45\times 10^{-7}$

Putting in the values we get:

$1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}$

$(\alpha)=0.0016$

$[H^+]=c\times \alpha$

$[H^+]=0.056\times 0.0016=8.96\times 10^{-5}$

Thus $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$

8 0

2 years ago

The mass of a solid substance is 21.112 g. If the volume of the solid substance is 19.5 cm3, calculate the density of the substa

Ostrovityanka [42]

Answer:

ρ = 1.08 g/cm³

Explanation:

Step 1: Given data

Mass of the substance (m): 21.112 g

Volume of the substance (V): 19.5 cm³

Step 2: Calculate the density of the substance

The density (ρ) of a substance is equal to its mass divided by its volume.

ρ = m / V

ρ = 21.112 g / 19.5 cm³

ρ = 1.08 g/cm³

The density of the substance is 1.08 g/cm³.

6 0

3 years ago

Part A

Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

7.33 × 10 ⁻³ c

Explanation:

1) Data:

a) m = 9.11 × 10⁻³¹ kg

b) λ = 3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

2) Formula:

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

λ = h / (m.s)

h is Planck's constant: h= 6.626×10⁻³⁴J.s

3) Solution:

Solve for s:

s = h / (m.λ)

Substitute:

s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s

s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

2.06 × 10 ⁻³⁴ m.

Explanation:

1) Data:

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ = ?

2) Formula:

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

λ = h / (m.s)

h is Planck's constant: h= 6.626×10⁻³⁴J.s

3) Solution:

λ = h / (m.s)

Substitute:

λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.

5 0

2 years ago

An aqueous solution containing 35.5 g of an unknown molecular (non-electrolyte) compound in 151.0 g of water was found to have a

Tasya [4]

ccccccccccccccccccccccccccccccccccc

8 0

3 years ago