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belka [17]
2 years ago
9

What is Cesium abbreviated

Chemistry
1 answer:
lubasha [3.4K]2 years ago
7 0
Celsius is °C , hope this helps
You might be interested in
What is the chemical equation
Serggg [28]

Chemical equation is the symbolic representation of chemical reactions.

Explanation

Chemical reactions are known as the reaction where two or more molecules or compounds react with each other leading to formation of product compounds along with either release or absorption of energy.

The symbolic representation of the processes occurring in a chemical reaction is termed as chemical equation.

The symbolic representation includes the reactants, products, external energy type and quantity of external energy and also about release of energy if occurs.

So the reactants are usually written in the left side of the chemical equation whereas on the right the products are written.

Both the sides are linked by a single headed arrow mark.

Some both the sides are linked by double heated arrow mark indicating the equilibrium chemical reaction.

5 0
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
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
11 months ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
Why do atoms gain or lose electrons?
Aleksandr [31]

Answer:kapil

Explanation:

Atoms lose electrons, if an electron gets more energy than then binding energy of the electron. This may be because of a collision with a charged particle or because of absorbtion of a photon. In a metal, there are just other positive charges nearby. The electron is not lost, but shared.

6 0
3 years ago
14 points!!! Pls tell me answer :3
iren [92.7K]
I’d say the last one
8 0
3 years ago
Read 2 more answers
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