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3 years ago

The formation of nitrogen dioxide and oxygen,

Chemistry

1 answer:

kvasek [131]3 years ago

7 0

Answer:

The rate of reaction changes by a factor 6.

Explanation:

Given the reaction is first order.

Let us assume the initial concentration of $NO$ is $A$ and the initial concentration of $O_3$ is $B$ .

So, we can write the rate of the reaction as

$R=k[A][B]$

Where,

$R$ is the rate of reaction. And $k$ is the rate constant.

Also, the initial concentration of $NO$ has been increased by $0.5$ , and the initial concentration of $O_3$ was increased by factor $4$ .

So, we have

$R'=k[1.5A][4B]\\R'=6k[A][B]\\R'=6R$

We can see the rate of reaction changes by a factor 6.

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A strip of copper 370 μm thick and 4.5 mm wide is placed in a uniform magnetic field B with arrow of magnitude 0.30 T, with B wi

777dan777 [17]

Explanation:

The given data is as follows.

Magnetic field (B) = 0.30 T

Current (I) = 24 A

n = $370 \times 10^{-6} m$

charge on electron (e) = $1.602 \times 10^{-19}$ C

Formula to calculate the potential difference is as follows.

V = $\frac{B \times I/d}{n \times e
}$

= $\frac{0.30 T \times 24 amps}{8.47 \times 10^{28} electrons/ m^{3} \times 370 \times 10^{-6} m \times 1.602 \times 10^{-19} C)}$

= $1.38 \times 10^{-6}$ V

Therefore, we can conclude that value of potential difference is $1.38 \times 10^{-6}$ V.

8 0

3 years ago

2) A professor wanted to set up a similar experiment as the one you performed in lab. He wanted to use Al(OH)3 in place of Ca(OH

expeople1 [14]

Explanation:

The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.

This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated

The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.

6 0

3 years ago

(CH3)2-CH-CH2-O(CH3)3IUPAC NAME

bazaltina [42]

Answer:

1-(tert-butoxy)-2-methylpropane

Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.

Explanation:

<em>Structural formula is attached</em>

IUPAC naming rules

1. start numbering the chain from the functional group. In this compound we start from oxygen side.

2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.

3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.

4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane

5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane

3 0

3 years ago

N=4 has how many orbitals?

marin [14]

Answer:

n = 4 there are 16 orbitals

5 0

2 years ago

A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete

lys-0071 [83]

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

8 0

3 years ago