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2 years ago

TIMED PLS HELP AND WILL GIVE BRAINLIST

Chemistry

1 answer:

Strike441 [17]2 years ago

6 0

Answer:

19.4 g of alum, will be its theoretical yield

Explanation:

The reaction is:

2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O

Let's determine the amount of acid.

M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution

M = mmol /mL

M . mL = mmol

We replace: 8.3 mL . 9.9 M = 82.17 mmoles

We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles

Ratio is 4:2

4 moles of sulfuric acid can make 2 moles of alum

By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.

We convert moles to mass:

Molar mass of alum is: 473.52 g/mol.

0.041085 moles . 473.52 g/mol = 19.4 g

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When air is heated, its density<br> A. Increases<br> B. Decreases<br> C. stays the same

soldier1979 [14.2K]

Answer:

Explanation:

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2 years ago

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The rate constant, k, for a reaction is 0.0354 sec1 at 40°C. Calculate the rate constant for the

deff fn [24]

Answer:

The rate constant of the reaction at 125˚ is $0.3115 \ \text{sec}^{-1}$ .

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

$k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}$

$\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}$

where $k$ is the rate constant, $E_{a}$ represents the activation energy of the chemical reaction, $R$ is the gas constant, $T$ is the temperature, and $A$ is the frequency factor.

The frequency factor, $A$ , is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be $k_{1}$ at an initial temperature $T_{1}$ and the rate constant $k_{2}$ at a final temperature $T_{2}$ , thus

$\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)$

$\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}$

Given that $E_{a} \ = \ 26.5 \ \ \text{kJ/mol}$ , $R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}$ , $T_{1} \ = \ \left(40 \ + \ 273\right) \ K$ , $T_{2} \ = \ \left(125 \ + \ 273\right) \ K$ , and $k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}$ , therefore,

$k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}$

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2 years ago

What is the boiling point of a solution of 12.0 g of urea in 165.0 g of water

Nat2105 [25]

Answer:

T = 100.63 °C

Explanation:

To solve this question, we need to know what are we talking about here. In this case, we want to know the boiling point of a solution with Urea in water. This is a colligative property, so, the expression to use to calculate that is the following:

ΔT = m * K / MM * kg water (1)

Where:

ΔT: difference of temperatures (Tb of solution - Tb water)

m: mass of the urea

K: ebulloscopic constant of the water (0.52 ° C / m)

MM: molecular mass of urea

The boiling point of water is 100 °C, we have the mass of the urea, but not the molar mass. The urea has the formula CH₄N₂O, so the molar mass can be calculated using the atomic mass of the elements (I will use a rounded number for this):

MM = 12 + (4*1) + (2*14) + 16 = 60 g/mol

Now, we can calculate the ΔT and then, the boiling point of the solution:

ΔT = 12 * 0.52 / 60 * 0.165

ΔT = 6.24 / 9.9

ΔT = 0.63 °C

the value of ΔT is a difference between the boling point of water and the solution so:

ΔT = Ts - Tw

Ts = ΔT + Tw

Replacing we have:

Ts = 100 + 0.63

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3 years ago

five atoms of hydrogen are reactants in a chemical reaction. how many atoms of hydrogen will be found in the products of that re

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Answer:

5 atoms of hydrogen will be found in the products

Explanation:

5 atoms of hydrogen will be found in the products becuase the law of conservation of mass states that mass is neither created nor destroyed in chemical reactions.

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2 years ago

Can you explain the difference between molarity, percent by mass, and molality?

cestrela7 [59]

Http://chemistry.about.com/od/chemistryterminology/a/What-Is-The-Difference-Between-Molarity-And-Mol...

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3 years ago