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Sergeeva-Olga [200]
3 years ago
15

Consider the reaction: 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) A suggested mechanism for this reaction follows: (1) NO(g) + NO(g) → N2

O4(g) (slow) (2) N2O4(g) + H2(g) → N2(g) + H2O2 (g) (fast) (3) H2O2(g) + H2(g) → 2H2O(g) (fast) Based on this mechanism, which, if any, of the following actions will not affect the rate of the reaction?
Chemistry
1 answer:
Dominik [7]3 years ago
4 0

Hello. This question is incomplete. The full question is:

"Consider the following reaction.  2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

A proposed reaction mechanism is:  NO(g) + NO(g) N2O2(g) fast  N2O2(g) + H2(g) → N2O(g) + H2O(g) slow  N2O(g) + H2(g) → N2(g) + H2O(g) fast

What is the rate expression?  A. rate = k[H2] [NO]2  B. rate = k[N2O2] [H2]  C. rate = k[NO]2 [H2]2  D. rate = k[NO]2 [N2O2]2 [H2]"

Answer:

A. rate = k[H2] [NO]2

Explanation:

A reaction mechanism is a term used to describe a set of phases that make up a chemical reaction. In these phases a detailed sequence of each step is shown, composed of several complementary reactions, which occur during a chemical reaction.

These mechanisms are directly related to chemical kinetics and allow changes in reaction rates to be observed in advance.

Reaction rate, on the other hand, refers to the speed at which chemical reactions occur.

Based on this, we can observe through the reaction mechanism shown in the question above, that the action "k [H2] [NO] 2" would have no changes in the reaction rate.

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Anna71 [15]
There’s 1024 atoms of sodium in 2 moles of sodium
7 0
3 years ago
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

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2 years ago
What is the osmotic pressure, in torr, of a 3.00% solution of NaCl in water when the temperature of the solution is 45 ºC? Enter
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213034 torr is the osmotic pressure.

Explanation:

osmotic pressure is calculated by the formula:

osmotic pressure= iCrT

where i= no. of solute

c= concentration in mol/litre

R= Universal Gas constant

T = temp

It is given that solution is 3% which is 3gms in 100 ml.

let us calculate the concentration in moles/litre

3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl

= 5.372 gm/litre

Putting the values in the formula,                   Temp in Kelvin 318.5K

osmotic pressure= 2*5.372*0.083 * 318.5                 Gas constant  0.083

                              = 284.023 bar or 213018 torr.                 c=  5.372 moles/L                                                                              

                                                                                    i=2 for NaCl

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D

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