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3 years ago

Identify the medium for the following waves:

Chemistry

1 answer:

Serjik [45]3 years ago

7 0

1. I don't understand what wave exercise waves are?

2. Water is the medium.

3. Air is the medium.

4. The Earth or ground is the medium.

(I think these are right, as the medium is what they are travelling through.)

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The name of an alcohol side chain is known as an alkoxy group. True or false?

mylen [45]

By itself, i don’t think so.

though, paired with a hydrogen bond, it is.

If i’m wrong, please feel free to let me know :)

6 0

3 years ago

Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea

REY [17]

Explanation:

Mg(s) + Cr(C2H3O2)3 (aq)

Overall, balanced molecular equation

Mg(s) + Cr(C2H3O2)3(aq) --> Mg(C2H3O2)3(aq) + Cr(s)

To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.

An increase in oxidation number denotes that the element has been oxidized.

A decrease in oxidation number denotes that the element has been reduced.

Oxidation number of Mg:

Reactant - 0

Product - +3

Oxidation number of Cr:

Reactant - +3

Product - 0

Note: C2H3O2 is actually acetate ion; CH3COO- The oxidatioon number of C, H and O do not change.

Oxidized : Mg

Reduced : Cr

7 0

3 years ago

Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100

vesna_86 [32]

<u>Answer:</u> The solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

$Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}$

3s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2$

We are given:

$K_{sp}=2.5\times 10^{-33}$

Putting values in above equation, we get:

$2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L$

Hence, the solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$

6 0

3 years ago

36g of an alloy of copper and zinc contains 45% copper. how much pure copper do you need to add in order to get a 60% copper all

iogann1982 [59]

First, you need to count copper mass in alloy.
Second, you have to make an equation an find x ( the copper mass must be added). The answer is: 13,5g pure copper

7 0

3 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago