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3 years ago

Count the atoms in 717 grams of Nitrogen.

Chemistry

1 answer:

MrRissso [65]3 years ago

3 0

Answer:

3.08x10^25

Explanation:

Divide the mass in grams by the molar mass of nitrogen (14.0067), then multiply the result by Avogadro's number (6.022x10^23). Then round to the correct amount of significant figures.

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A sample was prepared by mixing 18. ml of 3.00 x 10^-3 m crystal violet (cv) with 2.00 ml of 0.250 m naoh. calculate the resulti

Aleks [24]

Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

$Molarity = \frac{mol}{L}$

Molarity of crystal violet = $3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}$

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = $3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}$

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = $Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}$

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = $\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}$

Final concentration of CV = 0.0027 M

Molarity of NaOH= $\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}$

NaOH is a strong base and dissociates completely as follows.

$NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)$

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

8 0

3 years ago

Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol

zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = $10^{-4.8}$

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = $10^{-7.6}$

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = $10^{-1.9}$

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = $10^{1.8}$

[In-]/[HIn] = 63.10

6 0

3 years ago

HELP ME PLEASE ILL MARK U THE BRAINLIEST

ElenaW [278]

Answer:

pentene-alkanes-C5H10

octane-alkanes-C 8H 18

butene-alkene-c4h8

decane-alkane-c10h22

6 0

3 years ago

I NEED HELP PLEASEEEEEE I WILL RATE BRAINLIEST!!!!

MrRa [10]

<span>0.002 moles is the answer </span>

8 0

3 years ago

The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions

musickatia [10]

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,

x as 0.115 and y as 0.165

Now the mass of AgCl will be,

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,

278.1 × 0.165 = 45.88 grams.

3 0

3 years ago