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liberstina [14]
3 years ago
9

Balancing Chemical Equations & Identifying Chemical Reactions

Chemistry
1 answer:
Stells [14]3 years ago
3 0
The answer is 17 and a half
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Which statements are true about light waves? (Select all that apply.)
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A C D I believe

Explanation:

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The first step in using time more efficiently is
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The answer is studying how people use time.

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3 years ago
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16.0 grams of molecular oxygen gas is equal to
IceJOKER [234]

One thing to notice in the question is, we are asked about molecular oxygen that has formula O2 not atomic oxygen O.

As we are asked about molecular oxygen, we will answer the question in terms of number of molecules that are present in 16 grams of molecular oxygen.

To get the number of molecules present in 16 grams of O2, we will use the formula:

         No. of molecules = no. of moles x Avogadro's number (NA)-----  eq 1)

As we know:

                        The number of moles = mass/ molar mass of molecule

Here we have been given mass already, 16 grams and the molar mass of O2 is 32 grams.

Putting the values in above formula:

                                                    = 16/32  

                                                     = 0.5 moles

Putting the number of moles and Avogadro's number (6.02 * 10^23) in eq 1

                                No. of molecules = 0.5  x 6.02 * 10^23

                                   =3.01 x 10^23 molecules

or                 301,000,000,000,000,000,000,000 molecules

This means that 16 grams of 3.01 x 10^23 molecules of oxygen.

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4 0
3 years ago
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5 lists of physical change
Alecsey [184]
Crushing a can, breaking glass, cutting paper, boiling water, chopping wood, and mixing water and sand
6 0
3 years ago
What is the ph of a solution of 1.699 l of 1.25 m hcn, ka = 6.2 x 10-10, and 1.37 moles of nacn?
BlackZzzverrR [31]

The pH of a solution is 9.02.

c(HCN) = 1.25 M; concentration of the cyanide acid

n(NaCN) = 1.37 mol; amount of the salt

V = 1.699 l; volume of the solution

c(NaCN) = 1.37 mol ÷ 1.699 l

c(NaCN) = 0.806 M; concentration of the salt

Ka = 6.2 × 10⁻¹⁰; acid constant

pKa = -logKa

pKa = - log (6.2 × 10⁻¹⁰)

pKa = 9.21

Henderson–Hasselbalch equation for the buffer solution:

pH = pKa + log(cs/ck)

pH = pKa + log(cs/ck)

pH = 9.21 + log (0.806M/1.25M)

pH = 9.21 - 0.19

pH = 9.02; potential of hydrogen

More about buffer: brainly.com/question/4177791

#SPJ4

8 0
1 year ago
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