Acids react with

For a certain reaction, the frequency factor A is 5.0 × 109 s−1 and the activation energy is 16.3 kJ/mol. What is the rate const

WITCHER [35]

Answer:

K(79°C) = 1.906 E7 s-1

Explanation:

Arrhenius eq:

K(T) = A e∧(- Ea/RT)

∴ A = 5.0 E9 s-1

∴ Ea = 16.3 KJ/mol

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 79°C ≅ 352 K ⇒ K = ?

⇒ K(79°C) = (5.0 E9 s-1)e∧[ - (16.3KJ/mol)/(8.314 E-3 KJ/K.mol)(352 K)]

⇒ K(79°C) = (5.0 E9 s-1)e∧(- 5.5697)

⇒ K(79°C) = (5.0 E9 s-1)*(3.811 E-3)

⇒ K(79°C) = 1.906 E7 s-1

3 0

4 years ago

A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.9440-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

0.0133 mole = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.171g$

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%$

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

7 0

4 years ago

The first species that colonizes new or undisturbed land is called a _______ species. A. predecessor B. premier C. pioneer D. pr

Bond [772]

They would be called a pioneer species.

5 0

3 years ago

How are radioactive isotopes used to determine the absolute age of igneous rock?

Likurg_2 [28]

Answer:

d) all of the above

5 0

3 years ago

How many grams of dextrose are needed to make 725 mL of a 26.0% (w/v) dextrose solution? Note that mass is not technically the s

Diano4ka-milaya [45]

Answer:

188.5g of dextrose are needed

Explanation:

In Weight per volume percentage - %(w/v) -, the concentration is defined as the mass of solute in grams -In this case, dextrose-, in 100mL of solution.

As you want to prepare 725mL of a 26.0% (w/v) solution. you need:

725mL * (26g / 100mL) = 188.5g of solute =

<h3>188.5g of dextrose are needed</h3>

3 0

3 years ago