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frutty [35]
3 years ago
8

This is how fluorine appears in the periodic table. A green box has F at the center and 9 above. Below it says fluorine and belo

w that 19.00. A blue arrow points to 9. What information does "9” give about an atom of fluorine? Select three options. the atomic number the atomic mass the number of protons the number of electrons the number of neutrons
Chemistry
1 answer:
Oksana_A [137]3 years ago
8 0
1. Atomic weight of fluorine

2. Number of proton and neutrons found in the nucleus of the element

3. And every number uniquely identifies each of the the elements
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A sample of thionylchloride, SOCl2, contains 0.206 mol of the compound. What is the mass of the sample, in grams?
Ilia_Sergeevich [38]

Explanation:

Moles=mass/molar mass

moles × molar mass = mass

0.206 x 119= mass

Mass= 24. 51grams

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3 years ago
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Which of the following is a unit for volume?<br> m3<br> km<br> cg<br> dkm
babymother [125]
M³ is a unit for volume.
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Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of
Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

5 0
4 years ago
6.285×10^3 mg = _____? _____ kg
TiliK225 [7]
One kilogram is equal to one thousand grams. Further, one gram is equal to 1000 mg. The conversion is as shown below,
                    (6.285 x 10³ mg) x (1 g / 1000 mg) x (1 kg / 1000 g)
The numerical value of the operation above is 0.006285 kg. 
6 0
4 years ago
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Using wolfram alpha or some other reference, determine which of these elements would be liquid at 525 k (assume samples are prot
romanna [79]
The choices for this problem are bismuth, Bi; platinum, Pt; selenium, Se; calcium, Ca and copper, Cu. I think the correct answer would be selenium. The melting point of bismuth is at a temperature of 544.4 Kelvin. At a temperature of 525 K, it would exist as solid. Platinum melts at 2041.1 K. At 525 K, platinum would be in solid form. Selenium has a melting point at 494 K so that at a temperature of 525 K, it would exist in its liquid state. Calcium has a melting point of 1112 K so it would exist as solid at 525 K. Copper has a melting point at 1358 K, so it would still exist as solid at a temperature of 525 K. Therefore, the answer would only be selenium.
3 0
3 years ago
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