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Nikolay [14]
2 years ago
10

Which of the following is a direct result of global warming that is causing endangerment of polar bears?

Chemistry
1 answer:
frez [133]2 years ago
3 0

increased warmth causing glaciers to melt

you're welcome! :)

You might be interested in
Given the following thermodynamic data, calculate the lattice energy of LiCl:
tiny-mole [99]

Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

3 0
2 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
9966 [12]

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\%

5 0
1 year ago
how much hydrogen will be released during the interaction of potassium weighing 8 grams with hydrochloric acid
krek1111 [17]

Answer:

With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:

2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)

Explanation:

\large\colorbox{yellow}{ɪ ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs}

\large\colorbox{lightblue}{Xxᴊᴀsʜ13xX}

4 0
2 years ago
Repeating tests usually leads to
ANEK [815]

Answer:

It usually leads to more confidence in the results

4 0
3 years ago
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Peut-on être exact sans être précis?
Evgesh-ka [11]

Answer:

I think you can.

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