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algol [13]
3 years ago
8

For an ideal gas condition, what is the mass (g) of N2 if the pressure is 2.0 atm, the volume is 25 mL and the temperature is 29

0 Kelvin.
Chemistry
1 answer:
Alchen [17]3 years ago
3 0

Answer:

THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g

Explanation:

In an ideal condition

PV = nRT or PV = MRT/ MM where:

M = mass = unknown

MM =molar mass = 28 g/mol

P = pressure = 2 atm

V = volume = 25 mL = 0.025 L

R = gas constant = 0.082 L atm/mol K

T = temperature = 290 K

n = number of moles

The gas in the question is nitrogen gas

Molar mass of nitrogen gas = 14 * 2 = 28 g/mol

Then equating the variables and solving for M, we have

M = PV MM/ RT

M = 2 * 0.025 * 28 / 0.082 * 290

M = 1.4 / 23.78

M = 0.0589 g

The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g

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Answer: Finding the [H3O+] and pH of Strong and Weak Acid Solutions The larger the Ka, the stronger the acid and the higher the H+ concentration at equilibrium. hydronium ion, H3O+, 1.0, 0.00, H2O, 1.0×10−14, 14.00.

Explanation:The hydrogen ion in aqueous solution is no more than a proton, a bare ... the interaction between H+ and H2O .

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3 years ago
i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i
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First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

C_8H_{18}+O_2\to CO_2+H_2O

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

C_8H_{18}+O_2\to8CO_2+H_2O

Now, we balance the hydrogen:

C_8H_{18}+O_2\to8CO_2+9H_2O

And in the end, the oxygen:

C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O

We can multiply all coefficients by 2 to get integer ones:

2C_8H_{18}+25O_2\to16CO_2+18H_2O

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}

For the products, we have:

\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}

Now, we substract the rectants from the produtcs:

\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, <em>m</em>.

First, we have to heat the ice to 0 °C, so:

\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}

Then, we need to melt all this mass, so we use the latent heat now:

Q_2=n\cdot6.03kJ/mol

Converting mass to number of moles of water we have:

\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}

So:

Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g

Adding them, we have a total heat of:

\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}

Since we have a total of 20kg of ice, we can clculate the percent using it:

P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%

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