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IRISSAK [1]
4 years ago
11

14. A child's lungs can hold 2.20 L. How many grams of air do her lungs hold at a pressure of

Chemistry
2 answers:
Alenkasestr [34]4 years ago
5 0

Answer:

Explanation:

Use the gas equation: PV=nRT

P=pressure

V=Volume

R= gas constant of around 8.31 J/K/mol

T=temperature

n= number of moles

To find n, Rearrange:

n=PV/RT

102kPa= 102,000 kilo pascals which standard form is 102 x 10^3

Convert Celsius to kelvin, which you just add 273.15. So:

37+273.15=310.15 K round to a whole number is 310 K

Sub in all numbers to calculate the mol

n= 102 x 10^3 x 2.20 x 10^3/ 8.31 x 310 (cross out 10^3 as this will make a big number)

n=102 x 2.2/8.31 x 310 =0.087 mol

We know 1 g=29 moles

Multiply 29 moles with 0.087 to find the grams

29*0.087=2.523, which to one d.p is 2.5 g

Hence, the child's lung will hold 2.5 g of air.

Hope this helps you  :)

Have a nice day!!

Rom4ik [11]4 years ago
3 0

Answer:

The mass of air is 2.53 grams

Explanation:

Step 1: Data given

Volume of the lungs = 2.20 L

Pressure = 102 kPa = 1.00666 atm

Body temperature = 37.0 °C = 310 K

Molar mass of air = 29 g/mol

Step 2: Calculate number of moles

p*V = n*R*T

⇒with p = the pressure = 1.00666 atm

⇒with V = the volume of the lungs = 2.20 L

⇒with n = the number of moles air = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 310 K

n = (p*V)/(R*T)

n = (1.00666 * 2.20) / (0.08206* 310)

n = 0.0871 moles

Step 3: Calculate mass of air

Mass air = moles of air * molar mass

Mass air = 0.0871 moles * 29 g/mol

Mass air = 2.53 grams

The mass of air is 2.53 grams

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A 78.0 sample of gold at 175°C is put in a foam cup calorimeter with 25.0 g of water at 25.0°C. Calculate the specific heat of t
Nady [450]

The answer is the specific heat of the gold is 0.129 J/gC .

<h3>What is Calorimetry ?</h3>

It is an experimental method that allows one to calculate the heat change in a chemical process.

Calorimeter is just a reaction vessel. It could be a foam cup, a soda can, or a commercially available bomb calorimeter

Where q represents the heat change in the reaction, the calorimeter and the water. Since you are only measuring temperature, you will need to calculate the heat using it.

The change in heat of the gold is given by:

\rm q_{gold} = m c_{gold}\Delta T

=78 \times c_{p}\times (175-38.18)

The change in heat of the water is given by:

\rm q_{water} = mc_{p}\Delta T

m is the mass of water in the calorimeter in grams,

and delta T is the change in temperature.

where \rm c_{p} is the specific heat of water, which is 4.184 J/gC

mass of water =  25 g of water.

Temperature T₁ = 25°C

Final Temperature = 38.18 °C

Difference between temperature = 13.18 °C

\rm q_{water} = 25 \times 4.18\times 13.18

For a Foam Cup Calorimeter

\rm q_{water} = q_{cal}\\\\\\\rm 25\times4.18\times13.318 = 78 \times c_{p}\times (175-38.18)\\\\\\On \;solving\; this \;we \;get  \;c_{p} =  0.129

Therefore the  the specific heat of the gold is 0.129 J/gC

To know more about Calorimeter

brainly.com/question/4802333

#SPJ1

4 0
2 years ago
2 SO2(g) + O2(g) 2 SO3(g) Assume that Kc = 0.0680 for the gas phase reaction above. Calculate the corresponding value of Kp for
son4ous [18]

Answer: The corresponding value of K_p for this reaction at 84.5°C is 0.00232

Explanation:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

Relation of with is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

= equilibrium constant in terms of partial pressure = ?

K_c = equilibrium constant in terms of concentration = 0.0680

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature  =84.5^0C=(273+84.5)K=357.5K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

K_p=0.0680\times (0.0821\times 357.5)^{-1}\\\\K_p=0.00232

Thus the corresponding value of K_p for this reaction at 84.5°C is 0.00232

6 0
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