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3 years ago

A bumblebee flies with a ground speed of 15.0 m/s . Calculate its speed in km/h.

Chemistry

2 answers:

e-lub [12.9K]3 years ago

8 0

Answer:

$54\frac{km}{h}$

Explanation:

This is a problem of conversion of units.

First of all you need to take the quantity the problem gives you and write it down with the units:

$15.0\frac{m}{s}$

Then you need to find the conversion factor for the units you need to convert:

In this case you need to convert from meters (m) to kilometers (km) and from seconds (s) to hours (h), so you have the following conversion factors:

- To convert from meters (m) to kilometers (km):

$15.0\frac{m}{s}*\frac{1km}{1000m}$

-To convert from seconds (s) to hours (h):

$15.0\frac{m}{s}*\frac{3600s}{1h}$

Then you should put the conversion factors in the same expression, so:

$15.0\frac{m}{s}*\frac{1km}{1000m}*\frac{3600s}{1h}$

Finally you multiply the numerators and the denominators and divide the results:

$15.0\frac{m}{s}*\frac{1km}{1000m}*\frac{3600s}{1h}=54\frac{km}{h}$

zheka24 [161]3 years ago

5 0

Speed in km/hr = 15 x 18
------------
5

= 54 km/hr.

Hope this helps!

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3 years ago

A small portion of a crystal lattice is sketched below. What is the name of the unit cell of this lattice? Your answer must be a

Nikolay [14]

This is an incomplete question, the given sketch is shown below.

Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)

Explanation :

Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.

There are three types of unit cell.

SCC (simple-centered cubic unit cell)
BCC (body-centered cubic unit cell)
FCC (face-centered cubic unit cell)

In SCC, the atoms are arranged at the corners.

$Z=\frac{1}{8}\times 8=1$

The number of atoms of unit cell = Z = 1

In BCC, the atoms are arranged at the corners and the body center.

$Z=\frac{1}{8}\times 8+1=2$

The number of atoms of unit cell = Z = 2

The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.

$Z=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4$

The number of atoms of unit cell = Z = 4

Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)

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3 years ago

The addition of an inert gas has no effect on the equilibrium position of a gaseous reaction because ______. Multiple choice que

boyakko [2]

Inert gas does not affect the equilibrium position:

It is because the partial pressures of the reaction components remain the same.

What is Inert Gas?

Under a given set of conditions, an inert gas is a gas that does not undergo chemical reactions.
The noble gases (helium, neon, argon, krypton, xenon, and radon) were previously known as "inert gases" due to their perceived lack of involvement in any biochemical processes.
Because inert gases are non-reactive, they do not affect equilibrium partial pressures and thus do not affect volume.
An inert gas does not react with the reactants or products; it does not change the concentration of the products and reactants. Furthermore, because the volume is constant, the concentrations are unaffected. As a result, this does not affect equilibrium.

The equilibrium position won't change if an inert gas is added. A volume change won't change the equilibrium position if the total moles of gas in the products and reactants are the same. When the volume is reduced, the process changes to create fewer moles of gas.

Learn more about the inert gas here,

brainly.com/question/15909389

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2 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

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3 years ago

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sergejj [24]

Answer:

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Explanation:

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1 year ago