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3 years ago

A gas mixture has 10% O2, 50% Ar (40 gmw) and 40% Pu (244 gmw). What is the density of this mixture?

Physics

1 answer:

Irina-Kira [14]3 years ago

3 0

Answer: 5.39g/L

Explanation:

We need to calculate the average molar mass of the mixture which will be:

= (10 × 32) + (50 × 40) + (40 × 244) / 100

= (320 + 2000 + 9760) / 100

= 12080 / 100

= 120.8g/mol

Since the volume occupied is 22.4L and 1 mole of the gas will occupies 22.4L given the standard temperature conditions, then the density of the mixture will be:

= Mass / volume

= 120.8 / 22.4

= 5.39g/L

Therefore, the density of the mixture will be 5.39g/L.

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the answer is 47265.dug

Explanation:

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2 years ago

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emmainna [20.7K]

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3 years ago

An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring

lozanna [386]

Answer:

0.6 m

Explanation:

When a spring is compressed it stores potential energy. This energy is:

Ep = 1/2 * k * x^2

Being x the distance it compressed/stretched.

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The potential energy of the ice cube is:

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This is vertical height and is related to the distance up the slope by:

sin(a) = h/d

h = sin(a) * d

Replacing:

Ep = m * g * sin(a) * d

Equating both potential energies:

1/2 * k * x^2 = m * g * sin(a) * d

d = (1/2 * k * x^2) / (m * g * sin(a))

d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m

8 0

3 years ago

Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)

Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No

N = 0.1 No

Let the time taken is t and the decay constant is λ.

The relation between the decay constant and the half life is given by

$\lambda =\frac{0.6931}{T}=\frac{0.6931}{8}=0.08664 per day$

Us the equation of radioactivity

$N=N_{0}e^{-\lambda t}$

$0.1N_{0}=N_{0}e^{-0.08664 t}$

$e^{0.08664 t}=10$

Taking natural log on both the sides, we get

0.08664 t = 2.303

t = 26.6 days

4 0

3 years ago

Please answer D in the image with an explanation

puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

$T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}$

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

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2 years ago