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3 years ago

What is the purpose of the catalyst?

1 answer:

5 0

Its C

a catalyst speeds up a reaction by offering the reaction an alternative reaction pathway with a lower activation energy

hope that helps

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A gas occupies a volume of 584 mL at a pressure of 770 mm of Hg. When the pressure is changed, the volume becomes 603 mL If the

Blizzard [7]

Answer:

$P_2=795mmHg$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the final pressure by using the Boyle's law as an inversely proportional relationship in pressure to volume at constant temperature:

$\frac{P_2}{V_2} =\frac{P_1}{V_1}$

Thus, we solve for our target, P2, to obtain:

$P_2=\frac{P_1V_2}{V_1} =\frac{770mmHg*603mL}{584mL}\\\\P_2=795mmHg$

Regards!

4 0

3 years ago

The decantate from Question 1 that possibly contains K, Zn, or Cr, is neutralized with acetic acid and divided into the appropri

stepladder [879]

Answer:

2 that is Calcium (Ca) and Manganese (Mn)

Explanation:

Add dithizone in chloroform and look for a red complex formation.

4 0

4 years ago

Alcano con 1 carbono?

never [62]

creo esto te va a ayudar

5 0

3 years ago

Consider the exothermic reaction

alisha [4.7K]

Answer:

The answer to your question is -2855 J

Explanation:

Reaction

2C₂H₆ + 7O₂ ⇒ 4CO₂ + 6H₂O

Formula

Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants

Substitution

ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}

ΔHrxn = {-1574 -1450.8} - {-169.4}

ΔHrxn = -3024.8 + 169.4

ΔHrxn = -2855.4 J

4 0

3 years ago

Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.

Bezzdna [24]

Answer:

$\large \boxed{\text{ 0.066 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ: 58.44

NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g: 0.245

V/mL: 50.

c/mmol·mL⁻¹: 0.0180

2. Calculate the moles of each reactant

$\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}$

3. Identify the limiting reactant

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:

The molar ratio of NaCl to AgCl is 1:1.

$\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}$

From AgNO₃:

The molar ratio of AgNO₃ to AgCl is 1:1.

$\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}$

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)

I/mmol: 0.900 4.192 0

C/mmol: -0.900 -0.900 +0.900

E/mmol: 0 3.292 0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

$\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}$

8 0

3 years ago