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Nadusha1986 [10]
3 years ago
6

Total number of atoms in 4so2​

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

10

s=8

o=2

Hope it helps to

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An object has a density of 16.3 g/mL and a volume of 0.46 L. Calculate the
deff fn [24]

Answer:

16.53 pounds

Explanation:

this type of problem needs convertion method we need to convert from grams to pound.

3 0
3 years ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
3 years ago
7. How many moles are in 5 x 10 2 atoms of gold?
goblinko [34]

Answer:

8.3028894e-22

Explanation:

5x10^2 atoms/1 x 1 mol/6.022x10^23

7 0
3 years ago
Determine the average atomic mass of the following mixture of isotopes of Potassium:
dexar [7]

Answer

the first one (im pretty sure)

Explanation:

5 0
3 years ago
Find the % composition for each element in Zn(CIO3)2
bezimeni [28]

Zn = 28.15%

Cl = 30.53%

O = 41.32%

<h3>Further explanation</h3>

Given

Zn(CIO3)2 compound

Required

The % composition

Solution

Ar Zn = 65.38

Ar Cl = 35,453

Ar O = 15,999

MW Zn(CIO3)2 = 232.3

Zn = 65,38/232.3 x 100% = 28.15%

Cl = (2 x 35.453) / 232.3 x 100% = 30.53%

O = (6 x 15.999) / 232.3 x 100% = 41.32%

7 0
2 years ago
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