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gayaneshka [121]

2 years ago

11

How fast do you think the air particles in this room are moving compared to a car going 50 mph (about 22m/s)? Put your answer is

in the form, 'a molecule travels ___ as fast as a car"

1 answer:

Minchanka [31]2 years ago

5 0

The average molecule of air is moving at speed of 500 m /s and thus

speed of air particle = 500 m /s

speed of car = 22 m /s

Whereas, the speed of molecule /speed of car = 500 /22 = 22.72

Thus, molecule travel at 22.72 times faster than speed of car.

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$

(b) Since, the rate equation is as follows.

Rate = $K [OCl^{-}][l^{-}]$

Let us assume that ( $[OCl^{-}] = [l^{-}]$ )

Putting the given values into the above equation as follows.

$1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2$

$1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})$

K = $\frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}$

= $60.4 M^{-1}sec^{-1}$

Hence, the value of rate constant for the given reaction is $60.4 M^{-1}sec^{-1}$ .

(c) Now, we will calculate the rate as follows.

Rate = $K [OCl^{-}][l^{-}]$

= $60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})$

= $6.52 \times 10^{5}$

Therefore, rate when $[OCl^{-}] = 1.8 \times 10^{3}$ M and $[I^{-}]= 6.0 \times 10^{4}$ M is $6.52 \times 10^{5}$ .

8 0

2 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m15m. Assume that: The air pressure in his air tract is the same as the net wat

alexandr1967 [171]

Answer:

The ration of molar concentration is "2.5".

Explanation:

The given values are:

Average density of salt water,

= $1.03 \ g/cm^3$

Net pressure,

= $2.00 \ atm$

Increase in pressure,

= $1.00 \ atm$

Now,

The under water pressure will be:

= $\frac{15 \ m}{10}\times 1 \ atm +1 \ atm$

= $1.5\times 1+1$

= $1.5+1$

= $2.5 \ atm$

hence,

The ratio will be:

= $\frac{(\frac{n}{V})_{15m} }{(\frac{n}{V})_{surface} }$

or,

= $\frac{P}{P_s}$

= $\frac{2.5}{1}$

= $2.5$

7 0

2 years ago

How has Charles Darwin contributed to the study of evolution?

Lunna [17]

Do you answer is he proposed a genetic relationship between species

6 0

2 years ago

Help <br> whoever answers all gets marked brainliest

Lelechka [254]

#1

See H and C have shared their electrons so it's Covalent bonding

#2

Yes here we can see the dots and crosses clearly.

Dot and crosses digram

#3

It's Methane or CH_4

4 single bonds

#4

There is no double bond

6 0

2 years ago

Read 2 more answers

What is the name of the following chemical compound?

Lynna [10]

Answer: Dinitrogen pentoxide

Explanation:

5 0

2 years ago

Read 2 more answers