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kobusy [5.1K]
3 years ago
5

In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in

the solution slowly react with H2O(l) in the reversible reaction represented below.
BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

[HF] reaches a constant value of 0.0174 M when the reaction reaches equilibrium. For the forward reaction, the rate law is rate = kf [BF4-]. The value of the rate constant kf was experimentally determined to be 9.00x10-4 min-1.

Required:
a. Calculate the rate of the forward reaction after 600 minutes.
b. A student claims that the initial rate of the reverse reaction is equal to zero. Do you agree or disagree with this claim?
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
crimeas [40]

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0
3 years ago
How many moles of helium are needed to fill a balloon to a volume of 6.3 L at 28 °C and 320
Ray Of Light [21]

Answer:

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Explanation:

Convert mmHg to atms by dividing by 760. Then multiply 6.3 by the atms and divide by .08206*(273+28) to get mol

5 0
2 years ago
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LekaFEV [45]

The volume of a sample of ammonia gas : 5.152 L

<h3>Further explanation</h3>

Given

0.23 moles of ammonia

Required

The volume of a sample

Solution

Assumed on STP

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So for 0.23 moles :

= 0.23 x 22.4 L

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8 0
2 years ago
How many moles are present in 2.126 g of H2O2 ?
Vinil7 [7]

Answer:

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1 grams H2O2 is equal to 0.029399071224542 mole.

1 grams H2O2 to mol = 0.0294 mol

10 grams H2O2 to mol = 0.29399 mol

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30 grams H2O2 to mol = 0.88197 mol

40 grams H2O2 to mol = 1.17596 mol

50 grams H2O2 to mol = 1.46995 mol

100 grams H2O2 to mol = 2.93991 mol

200 grams H2O2 to mol = 5.87981 mol

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3 years ago
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lesantik [10]
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3 years ago
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