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TEA [102]
2 years ago
5

Define a function ComputeGasVolume that returns the volume of a gas given parameters pressure, temperature, and moles. Use the g

as equation PV = nRT, where P is pressure in Pascals, V is volume in cubic meters, n is number of moles, R is the gas constant 8.3144621 ( J / (mol*K)), and T is temperature in Kelvin.
Chemistry
1 answer:
andre [41]2 years ago
7 0

Answer: A function can be defined as a relation in which one thing is dependent on another for its value.

Explanation: Given R = 8.314J/mol*k

PV = nRT

V = nRT/ P

V = 8.314RT / P (cm^3)

Volume of gas =(( 8.314 * R* T) / P ) cm3

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serious [3.7K]
The reaction of HCl and NaOH is HCl + NaOH = NaCl + H2O. So the mole number of HCl and NaOH is equal. So the volume of HCl =0.01*0.1/0.02=0.05 L =50 ml. So the answer is D).
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Write a balanced equation for: (a) alpha decay of gold-173
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<span>The element gold has 32 isotopes, ranging from A =173 to A = 204. during alpha decay Au will loose 2 in atomic number and 4 in mass number . it will form a iridium isotope and helium . according to the above statement, the balanced equation for the alpha decay of gold 173 will be given as below. 173 169 4 79 Au-------------->77 Ir + 2He</span>
8 0
2 years ago
A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is
musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁  = 1.08atm

V₁  = 250mL

T₁  = 24°C

P₂  = 2.25atm

T₂  = 37.2°C

V₂  = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

              \frac{P_{1} V_{1} }{T_{1} }   =  \frac{P_{2} V_{2} }{T_{2} }

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

        T₁  = 24°C,  T₁   = 24 + 273 = 297K

        T₂  = 37.2°C , T₂  = 37.2 + 273  = 310.2K

Input the variables and solve for V₂

        \frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}

           V₂ = 124.91mL

6 0
2 years ago
mastering chem If the experiment below is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? At
Dvinal [7]

Answer: All of the statements are true.

Explanation:

(a) Considering the system mentioned in the equation:-

The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B

(b) 0<t< 20s,  mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.

0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,

40 to 60s is 0.14 (0.30-.16 = 0.14) mol

(0.46 > 0.14) mol leading this statement to be true as well.

(c) Average rate from t1 = 40 to t2 = 60 s is given by:

\delta moles/\delta time  = 0.30-0.16/60-40 = 0.007 Mol/s which is true as well

7 0
2 years ago
Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2
Svet_ta [14]

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

4 0
2 years ago
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