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wariber [46]
3 years ago
7

The feed to a batch process contains equimolar quantities of nitrogen and methane. write an expression for the kilograms of nitr

ogen in terms of the total moles n(mol) of this mixture.
Chemistry
1 answer:
timama [110]3 years ago
3 0
1) number of moles of N2 = n/2

2) Number of moles of CH4 = n/2

3) Total number of moles of the mixture = n/2 + n/2 = n

4) Kg of N2

mass in grams = number of moles * molar mass

molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol

=> mass of N2 in grams = (n/2) * 28 = 14n

mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg

Answer: mass of N2 in kg = 0.014n kg
You might be interested in
it is defined as looking for information by asking various questions about the thing you are curious about?
tino4ka555 [31]

Answer:

Inquiry.

Explanation:

Inquiry can be defined as the process (act) of looking for information by asking various questions about the thing you are curious about.

Simply stated, an inquiry is an act that typically involves seeking information by asking questions with respect to the thing you have curiosity for or the thing you wish to know more about.

This ultimately implies that, an inquiry deals with the urge to find information, seek for the truth or knowledge, examine principles and facts about the thing you are curious about.

3 0
3 years ago
What are the molarity and osmolarity of a 1-liter solution that contains half a mole of calcium chloride? How many molecules of
givi [52]

Answer:

Molarity = 0.5 M

Osmolarity = 0.5 x 2 = 1 Osmpl.

Molecules of Cl2 = 6.02 x 10^{23} / 4= 1.505 x 10^{23} no. of molecules

Explanation:

If we add half mole in 1L volume than molarity will obviously be 0.5 M.

The osmolarity is molarity multiplies by number of dissociates of solute that for CaCl2 are 2. So, 2 x 0.5 = 1

Half will be molecules of Ca and half will be of Cl2 for 0.5M.

6 0
4 years ago
A reaction has a rate constant of 2.08 × 10−4 s−1 at 26 oC and 0.394 s−1 at 79 oC . Determine the activation barrier for the rea
leva [86]

<u>Answer:</u> The activation energy of the reaction is 124.6 kJ/mol

<u>Explanation:</u>

To calculate activation energy of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{79^oC} = equilibrium constant at 79°C = 0.394s^{-1}

K_{26^oC} = equilibrium constant at 26°C = 2.08\times 10^{-4}s^{-1}

E_a = Activation energy of the reaction = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 26^oC=[26+273]K=299K

T_2 = final temperature = 79^oC=[79+273]K=352K

Putting values in above equation, we get:

\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol

Hence, the activation energy of the reaction is 124.6 kJ/mol

3 0
3 years ago
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

However; the workdone = -PdV

W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)

W = - 7.6 J

The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0
3 years ago
What is the empirical formula for a substance that contains 3.730% hydrogen, 44.44% carbon, and 51.83% nitrogen by mass?\?
Vlada [557]

Given data:

Hydrogen (H) = 3.730 % by mass

Carbon (C) = 44.44%

Nitrogen (N) = 51.83 %

This means that if  the sample weighs 100 g then:

Mass of H = 3.730 g

Mass of C = 44.44 g

Mass of N = 51.83 g

Now, calculate the # moles of each element:

# moles of H = 3.730 g/ 1 g.mole-1 = 3.730 moles

# moles of C = 44.44/12 = 3.703 moles

# moles of N = 51.83/14 = 3.702 moles

Divide by the lowest # moles:

H = 3.730/3.702  = 1

C = 3.703/3.702 = 1

N = 3.702/3.702 = 1

Empirical Formula = HCN

3 0
3 years ago
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