Ask question

Ask question

All categories

3 years ago

14

Which planets' orbits is the belt located between the main asteroid belt?

1 answer:

hichkok12 [17]3 years ago

3 0

<span>circumstellar is the answer</span>

You might be interested in

what's the main difference between continuous compounding and exponential growth, and why does my text book say that it is bette

Annette [7]

The big advantage to using continuous compounding to express growth rates is it avoids the problem of asymmetry in growth rates:

For example, if we use the normal definition and $100 grows to $105 in one time period, that's a growth rate of $105/$100 - 1 = 5% But if $105 decreases to $100, that's a growth rate of $100/$105 - 1 = -4.76%

The problem of asymmetry is those two growth rates, 5% and -4.75% are not equal up to a sign.

But if you use continuous compounding the growth rate in the first case is ln(105/100) = 0.04879.
And the growth rate in the second is ln (100/105) = -0.04879.
Those two growth rates are definitely the negative of each other.<span>
</span>

4 0

3 years ago

Which reaction takes place in a nuclear fission reactor?

Aleonysh [2.5K]

Answer:

It’s C I think Explanation:

7 0

3 years ago

Read 2 more answers

The u.S. Mint produced pure silver quarter commemorative coins in 1992, each coin weighing 5.67 g. How many silver atoms are pre

PolarNik [594]

Mass of each silver coin = 5.67 g

Molar mass of silver = 107.87 g/mol

Calculating the moles of silver from given mass and molar mass of silver:

$5.67 g Ag*\frac{1molAg}{107.87g} =0.0526molAg$

Calculating the atoms of silver in 0.05260 mol using the conversion factor, $1mol=6.022*10^{23}atoms$ :

$0.05260 mol*\frac{6.022*10^{23}atoms }{1mol} =3.17*10^{22}atoms$

Therefore, $3.17*10^{22}$ atoms of silver are present in each coin.

4 0

3 years ago

Starting with calcium chloride describe how one can prepare calcium carbonate

Dima020 [189]

I have no idea I’m sorry I’m doing a test and need to put a question

8 0

3 years ago

Consider the concentration cell in which the metal ion has a charge of +2, and the solution concentrations are: dilute solution

tia_tia [17]

<u>Answer:</u> The predicted cell potential of the cell is +0.0587 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u> $M(s)\rightarrow M^{2+}+2e^-$

<u>Reduction half reaction (cathode):</u> $M^{2+}+2e^-\rightarrow M(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[M^{2+}_{(diluted)}]}{[M^{2+}_{(concentrated)}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[M^{2+}_{(diluted)}]$ = 0.05 M

$[Zn^{2+}_{(concentrated)}]$ = 4.808 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{0.05M}{4.808M}$

$E_{cell}=0.0587V$

Hence, the predicted cell potential of the cell is +0.0587 V

4 0

3 years ago