All categories

3 years ago

Atmosphere pressure is about 100000 Pa. Calculate the force of the Atmosphere on a person whose Surface area is 2.0m

Chemistry

1 answer:

Sedaia [141]3 years ago

7 0

Answer:

The force of the atmosphere 200 KN.

Explanation:

Pressure is the ratio of force applied to the area to which the force is applied.

i.e Pressure = $\frac{Force}{Area}$

But, Pressure = 100000 pa (100000 N/ $m^{2}$ ), and surface area = 2.0 $m^{2}$

Thus,

100000 = $\frac{Force}{2.0}$

⇒ Force = 100000 x 2.0

= 200000

Force = 200 KN

The force of the atmosphere 200 KN.

You might be interested in

Joe is concerned about the carbon dioxide emissions from his daily commute and the impact that it is having on global warming In

Vesna [10]

Amount of CO₂ emission per day is 11,356.23 g.

Explanation:

Joe travelling distance per day = 60 miles

Carbon dioxide emission per day = 20 mpg

Now we have to find the amount of carbon dioxide emitted per day by dividing the distance by the emission per day given in gallons.

Amount of Carbon dioxide emission = $$\frac{distance}{emission amount}$

Amount of CO₂ emission in gallons = $$\frac{ 60 miles}{20 mpg}$

= 3 gallons

Now we have to convert the gallons to grams as,

1 gallon = 3,785.41 g

3 gallons = 3 × 3785.41 g = 11,356.23 g

So the emission of CO₂ per day is 11,356.23 g.

6 0

3 years ago

Ammonia (NH3) is widely used as a fertilizer

ratelena [41]

Answer:

82.28g

Explanation:

Given parameters:

Number of moles of hydrogen gas = 7.26 moles

Unknown:

Amount of ammonia produced = ?

Solution:

We have to write the balanced equation first.

N₂ + 3H₂ → 2NH₃

Now, we work from the known to the unknown;

3 moles of H₂ will produce 2 moles of NH₃

7.26 mole of H₂ will produce $\frac{7.26 x 2}{3}$ = 4.84 moles of NH₃

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Mass of NH₃ = number of moles x molar mass = 4.84 x 17 = 82.28g

3 0

3 years ago

A crane lifts a 5,800-N block from the ground to 20 m above the ground in 80 seconds. How much Power

earnstyle [38]

Answer:

1450 W

Explanation:

5800n x 20m =1450w

80s

3 0

3 years ago

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

To what temperature must a balloon, initially at 25°c and 2.00 l, be heated in order to have a volume of 6.00 l? question 8 opti

cluponka [151]

a.655 k not 100 percent on this but try it. You will use 273.15 and add your Celcius temp to get it in Kelvin

5 0

3 years ago

Read 2 more answers