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babymother [125]

3 years ago

9

Why does a mechanical wave require a medium for transmission?

1 answer:

Ganezh [65]3 years ago

6 0

The medium provides an opposing force to slow down the wave.

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The penguin has a ___ body that helps it to move quickly in water.

Lerok [7]

The penguin has a sleek body that helps it to move quickly in water.

7 0

3 years ago

Read 2 more answers

What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor

ddd [48]

Answer:

$1.75272\ \text{m}^3$

Explanation:

The breakdown reaction of ozone is as follows

$CF_3Cl + UV \rightarrow CF_3 + Cl$

$Cl + O_3 \rightarrow ClO + O_2$

$O_3 + UV \rightarrow O_2 + O$

$ClO + O \rightarrow Cl + O_2$

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of $CF_3Cl$ = 15.5 g

M = Molar mass of $CF_3Cl$ = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = $62.363\ \text{L mmHg/K mol}$

Number of moles is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}$

$20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}$

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3$

For 20 cycles of the reaction the volume of the ozone is $1.75272\ \text{m}^3$ .

7 0

3 years ago

How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?

Bad White [126]

The balanced chemical reaction is :

$O_2 + 4Na \ -> \ 2Na_2O$

Number of moles of Na, $n = \dfrac{14.6}{23} = 0.635 \ mol$ .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

$n = \dfrac{0.635}{4}\ mole$

So, amount of oxygen required is :

$m = \dfrac{0.635 \times 32}{4}\ gm\\\\m = 5.08 \ gm$

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0

2 years ago

What product has almost the same compounds as baking soda 10pts

mamaluj [8]

Ummm i actually dont know lol i swear i knew im sorry man or women ( im not trying to assume your gender

4 0

3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

<u>Answer:</u> The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u> $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

<u>Equation 2:</u> $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

5 0

3 years ago