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4 years ago

Decreasing the temperature of a reaction decreases the reaction rate because

Chemistry

2 answers:

nadezda [96]4 years ago

7 0

Answer: D) decreasing the temperature lowers the average kinetic energy of the reactants.

Explanation:

Rate of a reaction decreases on decreasing the temperature because the energy of the reactant decreases and thus less molecules can cross the energy barrier.

Average kinetic energy is defined as the average of the kinetic energies of all the particles present in a system. It is determined by the equation:

$K=\frac{3RT}{2}$

From above, it is visible that kinetic energy is directly related to the temperature of the system. So, if temperature is less, average kinetic energy of the system is less and thus rate decreases.

Elza [17]4 years ago

4 0

The answer u are looking for is b

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The thermometer in the picture reads, "19.9 degrees Celsius".

GalinKa [24]

Answer:

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Explanation:

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3 years ago

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Ulleksa [173]

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4 years ago

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Which of these describes how traits are shown on a branching tree diagram?

Sophie [7]

Answer:

D- Traits higher on the tree developed before traits lower on the tree.

Explanation:

The answer is D because the higher/stronger a trait is it will be developed before any lower traits.

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4 years ago

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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

A gas has an initial volume of 3.0 L with a pressure of 0.62 atm. What will be the new pressure if the volume changed to 1.0 L?

irina1246 [14]

Answer:

The new pressure will be 1.86 atm.

Explanation:

As the volume increases, the gas particles (atoms or molecules) take longer to reach the walls of the container and therefore collide less times per unit time against them. This means that the pressure will be less because it represents the frequency of gas strikes against the walls. In this way pressure and volume are related, determining Boyle's law.

Boyle's law says that at constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts:

P*V=k

To determine the change in pressure or pressure during a transformation to constant pressure, the following expression is applied:

P1*V1=P2*V2

In this case:

P1= 0.62 atm
V1= 3 L
P2= ?
V2= 1 L

Replacing:

0.62 atm* 3L= P2* 1 L

Solving:

$P2=\frac{0.62 atm*3L}{1L}$

P2= 1.86 atm

<u><em>The new pressure will be 1.86 atm.</em></u>

7 0

3 years ago