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NeTakaya
3 years ago
10

Which of these time measurements is the smallest?

Chemistry
2 answers:
TiliK225 [7]3 years ago
8 0
25,000 nanoseconds is the smallest time measurement
Anuta_ua [19.1K]3 years ago
7 0

Answer:

In order to answer the question, we convert the measurements given to a common base unit. For this case, we use seconds.

A)   0.02 seconds

B)   0.02 teraseconds x (10^12 s / 1 terasecond) = 2x10^10 seconds

C)   2,500 milliseconds x (1 s / 1000 ms) = 2.5 seconds

D)   25,000 nanoseconds x (1 x 10^-9 s / 1 nanosecond) = 2.5 x 10^-5 seconds

So ,The correct answer is option " D "

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Compound A has molecular formula C5H11Br. When compound A is treated with bromine in the presence of UV light, the major product
damaskus [11]

The compound (A) is 2-bromo-pentane. See attached picture for the chemical diagram.

Explanation:

2-bromo-pentane (A) reacts with bromine (Br₂) under UV light to form 2,2-dibromopentane and HBr. Also 2-bromo-pentane (A) reacts with NaSH to from pentane-2-thiol, which have an asimetric carbon, and NaBr.

You may find the chemical reactions and diagrams of the compounds in the attached picture.

Learn more about:

structure of organic compounds

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#learnwithBrainly

6 0
2 years ago
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wave
timofeeve [1]

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

n_i=8

\lambda=3745\ nm

The amount of energy change during the transition is given by :

\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]

And

\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]

Plugging all the values we get :

\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5

So, the final level of the electron is 5.

4 0
2 years ago
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
Anna71 [15]

Answer:

31.652g of Na3PO4

Explanation:

We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:

Na3PO4 will dessicate in solution as follow:

Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.

Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e

xM Na3PO4 = (1.10 x 1)/3

xM Na3PO4 = 0.367M

Therefore, the molarity of Na3PO4 is 0.367M.

Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:

Molarity of Na3PO4 = 0.367M

Volume = 525mL = 525/1000 = 0.525L

Mole of Na3PO4 =..?

Molarity = mole /Volume

0.367 = mole /0.525

Cross multiply

Mole of Na3PO4 = 0.367 x 0.525

Mole of Na3PO4 = 0.193 mole.

Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:

Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol

Mole of Na3PO4 = 0.193 mole

Mass of Na3PO4 =.?

Mass = mole x molar mass

Mass of Na3PO4 = 0.193 x 164

Mass of Na3PO4 = 31.652g

Therefore, 31.652g of Na3PO4 is needed to prepare the solution.

6 0
2 years ago
A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod
sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

brainly.com/question/27891343

#SPJ4

7 0
1 year ago
When there are more electrons than protons an object has an overall<br> charge.
azamat

Answer:it would have a negative charge because  electrons are negative

Explanation:

7 0
3 years ago
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