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3 years ago

A first order reaction has a rate constant of 0.543 at 25 c and 6.47 at 47

Chemistry

1 answer:

alukav5142 [94]3 years ago

5 0

The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

where :

k1 is the rate constant at temperature T1

k2 is the rate constant at temperature T2

R = gas constant = 8.314 J/K-mol

Given data:

k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K

k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K

ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]

2.478 = 2.774 *10^-5 Ea

Ea = 0.8934*10^5 J = 89.3 kJ

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Question 1<br> How many grams are equal to 3.4 moles of carbon dioxide?

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Answer:

149.6 grams

Explanation:

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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

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

7 0

3 years ago

The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p

almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen 40300 g

Carbon 11160 g

Hydrogen 6200 g

Nitrogen 1860 g

Calcium 992 g

Phosphorus 744 g

Other elements 744 g

It is possible to check it adding all the components, the total is 62000 g.

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

Percentage ( %)

Oxygen 65.0

Carbon 18.0

Hydrogen 10.0

Nitrogen 3.0

Calcium 1.6

Phosphorus 1.2

Other elements 1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

Percentage ( %) Decimal number

Oxygen 65.0 65/100 = 0.65

Carbon 18.0 18/100 = 0.18

Hydrogen 10.0 10/100 = 0.10

Nitrogen 3.0 3/100 = 0.03

Calcium 1.6 1.6/100 = 0.016

Phosphorus 1.2 1.2/100 = 0.012

Other elements 1.2 1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

Percentage ( %) Decimal number Mass in grams (g)

Oxygen 65.0 65/100 = 0.65 0.65*62000=40300

Carbon 18.0 18/100 = 0.18 0.18*62000= 11160

Hydrogen 10.0 10/100 = 0.10 0.10*62000= 6200

Nitrogen 3.0 3/100 = 0.03 0.03*62000= 1860

Calcium 1.6 1.6/100 = 0.016 0.016*62000= 992

Phosphorus 1.2 1.2/100 = 0.012 0.012*62000=744

Other elements 1.2 1.2/100 = 0.012 0.012*62000=744

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3 years ago

Exactly one mole of an ideal gas is contained in a 2.00 liter container at 1,000 K. What is the pressure exerted by this gas?

Kipish [7]

pV = nRT

p = nRT/V

p= 1 x 0.08205 x 1000/ 2

p = 41.025 Pa

Edit: The unit should be atm instead of Pa, as pointed out by a nice human being.

4 0

3 years ago

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