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ExtremeBDS [4]
3 years ago
13

Balance this equation and pick the answer with the coefficients that belong there.

Chemistry
1 answer:
geniusboy [140]3 years ago
6 0

Answer:

2,3,6,1

2,3,6,1

Explanation:

The unbalanced reaction expression is given as:

  AlBr₃    +    K₂SO₄  →    KBr  + Al₂(SO₄)₃

We need to balanced this reaction equation. Our approach is a mathematical method where we assign variable a,b,c and d as the coefficients.

 aAlBr₃    +    bK₂SO₄  →    cKBr  + dAl₂(SO₄)₃

Conserving Al;  a  = 2d

                   Br:  3a  = c

                   K:   2b = c

                   S:   b  = 3d

                   O:  4b  = 12d

Let a  = 1, c = 3, d  = \frac{1}{2}  b  = \frac{3}{2}  

 Multiply through by 2 to give;

     a  = 2, b = 3, c = 6 and d  = 1

 2AlBr₃    +    3K₂SO₄  →    6KBr  + Al₂(SO₄)₃

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3 years ago
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A philosophy of life which, by rejecting the democratic concept of the mass-man, endeavors to give this earth to the best nation
Julli [10]

Answer:

The correct answer is "a quote from Mein Kampf written by Adolf Hitler".

Explanation:

Mein Kampf is a 1925 famous book written by Adolf Hitler that serves as an autobiographical manifesto of the Nazi Party that he led. The statement that is included in this paragraph belongs to a quote of this book. In this statement it is reflected the ideology of Adolf Hitler, denying democracy as a government and expressing its ideology of a totalitarian autocracy.

5 0
3 years ago
N2(g) + 3H2(g) -&gt; 2NH 3(g)
Nutka1998 [239]

Answer:

None of the options are correct. The correct answer is:

56.67g

Explanation:

N2 + 3H2 —> 2NH3

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

From the equation,

6g of H2 produced 34g of NH3.

Therefore, 10g of H2 will produce = (10 x 34)/6 = 56.67g of NH3

Therefore, 56.67g of NH3 are produced

5 0
3 years ago
A 475 cm3 sample of gas at standard temperature and pressure is allowed to expand until it occupies a
Andrej [43]

The final temperature : 345 K

<h3> Further explanation </h3>

Given

475 cm³ initial volume

600 cm³ final volume

Required

The final temperature

Solution

At standard temperature and pressure , T = 273 K and 1 atm

Charles's Law  :

When the gas pressure is kept constant, the gas volume is proportional to the temperature  

V₁/T₁=V₂/T₂

Input the value :

T₂=(V₂T₁)/V₁

T₂=(600 x 273)/475

T₂=345 K

4 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
1 year ago
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