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Gennadij [26K]
3 years ago
15

When a 19.2-g sample of KCN dissolves in 65.0 g of water in a calorimeter, the temperature drops from 28.1 (C) to 15.4 (C). Calc

ulate DH for the process
Mathematics
1 answer:
Greeley [361]3 years ago
8 0

Answer:

The answer is "\bold{\ 15.4  \frac{kj}{mol}} \\ ".

Step-by-step explanation:

Equation:

KCN(s) \rightarrow K^+(aq)+CN^-(aq)    \ \ \ \  ΔH =?

\ Q = (84.2) (4.184)(-12.7) \\\\\ Q = -4474\\\\\ Q_(r \times r) = \frac{+ 4.474 kj}{0.29mol} \\\\\ Q= 15.4 \frac{kj}{mol}

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So, the ration is 17:3, you know that. Therefore, the change from # of students to the 17 is the same change as # teachers to 3. Knowing this, you also know that the total # of people ihas the same change as the total ratio. So 17+3 = 20. 740/20 = 37. So the change is 37. 3x37= 111   17x37=629 
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Solve the system of equations.
emmainna [20.7K]

x = \frac{ 419 }{ 113 } ~,~y = -\frac{ 208 }{ 113 } ~,~z = \frac{ 21 }{ 113 }

<h2>Explanation:</h2>

We have the following system of three linear equations:

\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\5~ x&+~~8~ y&+~~~~ z&~=~4\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}

Let's use elimination method in order to get the solution of this system of equation, so let's solve this step by step.

Step 1: Multiply first equation by -5/2 and add the result to the second equation. So we get:

\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}

Step 2: Multiply first equation by −2 and add the result to the third equation. So we get:

\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&-~~~3~ y&-~~~62~ z&~=~-6\end{array}

Step 3: Multiply second equation by −32 and add the result to the third equation. So we get:

\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&&+~~\frac{ 113 }{ 2 }~ z&~=~\frac{ 21 }{ 2 }\end{array}

Step 4: solve for z.

\begin{aligned}       \frac{ 113 }{ 2 } ~ z & = \frac{ 21 }{ 2 } \\      z & = \frac{ 21 }{ 113 }       \end{aligned}

Step 5: solve for y.

\begin{aligned}-2y-79z &= -11\\-2y-79\cdot \frac{ 21 }{ 113 } &= -11\\y &= -\frac{ 208 }{ 113 } \end{aligned}

Step 6: solve for x by substituting y=-\frac{208}{113} and z = \frac{21}{113} into the first equation:

2x+4(-\frac{208}{113})+32(\frac{21}{113})=6 \\ \\ 2x-\frac{832}{113}+\frac{672}{113}=6 \\ \\ 2x=6+\frac{832}{113}-\frac{672}{113} \\ \\ 2x=\frac{838}{113} \\ \\ x=\frac{319}{113}

Finally:

x = \frac{ 419 }{ 113 } ~,~y = -\frac{ 208 }{ 113 } ~,~z = \frac{ 21 }{ 113 }

<h2>Learn more:</h2>

Solving System of Equations: brainly.com/question/13121177

#LearnWithBrainly

7 0
4 years ago
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