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DochEvi [55]
3 years ago
9

Co (5. 00 g) and co2 (5. 00 g) were placed in a 750. 0 ml container at 50. 0 °c. The partial pressure of co in the container was

________ atm.
Physics
1 answer:
konstantin123 [22]3 years ago
6 0
Use PV = nRT and solve for P.
n = grams/molar mass = 5.00/molar mass CO2
Ignore the SO2.

But I think the answer it 6.74 atm

But I’m not saying but is but I’m just thinking this is the answer.

Goodluck!
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3 years ago
When a hailstone is at a height of 2.00km it’s mass is 2.50g what is it’s potential energy?
scZoUnD [109]

Answer:

EP = 49.05Joules (J)

Explanation:

The equation for Potential energy (EP) is

EP = m g h

We are given the values below (do convert them into SI units)

m = 0.0025kg

h = 2000m

g = 9.81m/s^{2}

Substitute the values into the equation and solve for EP

EP = 0.0025 * 2000 * 9.81

EP = 49.05Joules (J)

6 0
3 years ago
A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th
vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 (1)

where

p_1 = 7340 N/m^2 is the pressure in the pipe

p_2 = 5505 N/m^2 is the pressure in the constricted section

\rho = 821 kg/m^3 is the density of the oil

v_1 is the velocity of the oil in the pipe

v_2 is the velocity of the oil in the constricted section

Also, according to the continuity equation,

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the pipe, with

r_1 = \frac{1.03}{2}=0.515 m is the radius

A_2 = \pi r_2^2 is the cross-sectional area of the constricted section, with

r_2=\frac{0.618}{2}=0.309 m is the radius

So the equation becomes

r_1^2 v_1 = r_2^2 v_2

So we can write

v_2=\frac{r_1^2}{r_2^2}v_1

Substituting into eq.(1),

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2

And solving the equation for v_1:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s

And the velocity in the constricted section is

v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s

Learn more about flow rate:

brainly.com/question/9805263

#LearnwithBrainly

7 0
3 years ago
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