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Firdavs [7]
3 years ago
12

A certain reaction has the form aA → bB. At a particular temperature and [A]0 = 2.80 × 10-3 molar, data was collected of concent

ration versus time for this reaction. A plot of 1/[A]t versus time resulted in a straight line with a slope value of 3.60 × 10-2 M-1s-1. What is the concentration of A after 300 seconds?
a.2.80 × 10-3 molar
b.2.72 × 10-3 molar
c.2.60 × 10-3 molar
d.3.60 × 10-2 molar
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
7 0
If the plot of 1/[A] versus time shows a constant slope, this means it is a second-order reaction. The equation would be:
1/[A] - 1/[A]0 = kt
1/[A] - 1/0.0028 = 0.036(300)
1/[A] = 0.00272 = 2.72 x 10^-3 M
This is equivalent to choice B.
You might be interested in
A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this
NeTakaya

Answer:

a)  molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane CH_4}= 16 g/mol

Molar mass of ethane C_2H_6= 30 g/mol

Molar mass of ethylene C_2H_4 = 28 g/mol

Molar mass of water H_2O=18g/mol

number of moles = \frac{mass}{molar mass}

Their molar composition can be calculated as follows:

n_{CH_4}= \frac{75}{16}

n_{CH_4}= 4.69 moles

n_{C_2H_6} = \frac{10}{30}

n_{C_2H_6} = 0.33 moles

n_{C_2H_4} = \frac{5}{28}

n_{C_2H_4} = 0.18 moles

n_{H_2O}= \frac{10}{18}

n_{H_2O}= 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = \frac{n_{H_2O}}{n_{drygas}}

= \frac{0.56}{5.2}

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

    CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O

4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

    C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O

0.18           3× 0.18

moles        moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h

Total moles of O_2 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles O_2

Thus, moles of air required = \frac{1}{0.21}*11.075

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0
3 years ago
What type of elements are most likely to form more than one type of ion
ss7ja [257]
Metals are the type of elements that are most likely to form more than one type of ion, for instance iron can form the ion of Fe^2+ or Fe^3+.
7 0
3 years ago
Read 2 more answers
The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o
meriva

Answer:

\boxed{-267.5}

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}

The equation for the reaction is

                        C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹   219.5      205.0         213.6         69.9

\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}

3 0
3 years ago
URGENT! Consider the following decomposition reaction of ammonium carbonate ((NH4)2CO3):
dexar [7]
<span>Using PV=nRT to find the moles and then convert back.
</span><span>4x=.8944
</span><span>solve for x then use the pressure for lets say CO2 put that into PV=nRT then solve for n then convert over.
</span>
<span>(.2236)(2)/(298*.08206) = .0183*96g/mol = 1.76g
</span>
<span>For C:

[NH3]^2[CO2][H2O] = Kp
x=0.2236 (2*.2236)^2(.2236)*(.2236)
  =0.001
</span>
6 0
3 years ago
A sample of a substance that has a density of 0.824 g/mL has a mass of 0.451g. Calculate the volume of the sample.
Allushta [10]
Density can be calculated using the following rule:
density=mass/volume
therefore,
volume=mass/density

we have mass=0.451g and density=0.824g/ml
substituting in the above equation, we can calculate the volume as follows:
volume = 0.451/0.824 = 0.547 ml
8 0
3 years ago
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