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Fittoniya [83]
3 years ago
15

given the following hypothetical reaction 2E(g)+ F(g)+O(g)_products.if the rate law is rate _K(e)^2(f)^-1,what is the order of t

he reaction with respect to water?​
Chemistry
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer:

See explanation.

Explanation:

Hello there!

In this case, considering the given information, it is not possible to figure out such order of reaction with respect to water as it is not involved in the given chemical reaction:

2E(g)+ F(g)+O(g)_

However, we can say that the order of reaction is 2 with respect to E (power of 2 in the rate law), 1 with respect to F (power of 1 in the rate law) and 0 with respect to O (power of 0 or not present in the rate law).

Regards!

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Andreas93 [3]

There are 60 seconds in one minute

565900 / 60 = 9431.667 minutes in 565,900 seconds

There are 60 minutes in an hour

9431.667 minutes / 60 = 157.9144 hours in 565900 seconds

There are 24 hours in 1 day

157.9144 hours / 24 hours = 6.5498 days.

4 0
3 years ago
When KNO3 dissolved in water, what is the intermolecular attraction between NO3 ions
Snezhnost [94]
Ion-dipole forces

H2O has hydrogen bonding, which is a form of dipole-dipole forces, and NO3- is an ion, so the intermolecular attraction is ion-dipole.
4 0
3 years ago
Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. dete
oee [108]
Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
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5 0
3 years ago
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
4 years ago
What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7
mariarad [96]

 The  empirical  formula    of hydrocarbon is  CH2

The  molecular formula  of the  hydrocarbon is  C6H12


    <u><em>Explanation</em></u>

Hydrocarbon  is  made up  of carbon and hydrogen


<h3><u><em> </em></u>Empirical formula  calculation</h3>

 Step 1:  find  the  moles   CO2  and  H2O

moles =mass/molar mass

moles   of CO2 =  6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol =  0.15  moles

Step 2: Find the moles  ratio  of Co2:H2O  by diving  each mole by smallest mole(0.15)

that  is  for  CO2 = 0.15/0.15  =1

              For H2O = 0.15/0.15 =1

therefore  the mole ratio  of Co2 : H2O = 1:1  which  implies that 1 mole of Co2  and 1  mole of H2O is  formed  during combustion reaction.


From the  the law of mass conservation the number  of atoms in reactant side  must  be equal to  number of  atoms  in product side

therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

From information above the empirical formula is therefore = CH2


Molecular formula  calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n =  84 g/mol

n = 6

multiply the  each subscript  in CH2  by  6

 Therefore the molecular formula = C6H12




5 0
3 years ago
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