40 POINTS write a lab report about types of reactions

Chemistry

2 answers:

Masteriza [31]3 years ago

6 0

Answer:

The purpose of this lab was to explore the different types of chemical reactions in this laboratory procedure. The question for this lab is, how does knowing the reactants and products help me classify a chemical reaction. In this experiment, I made observations of the reactants and the products. I used these observations to write the balanced chemical equations and to classify the reaction it made.

Explanation:

frozen [14]3 years ago

5 0

Answer:

bruh

Explanation:

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VikaD [51]

Answer: 3 and 4 are the answers.

Explanation: i took the quiz.

7 0

2 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$