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mamaluj [8]
3 years ago
8

What is the total number of moles contained in 115g of C2H5OH?

Chemistry
1 answer:
julsineya [31]3 years ago
6 0
2.4962859606272754 moles of C2H5OH
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What are some elements that can exist in room temperature?
SOVA2 [1]
Most chemical elements are solid at room temperature. Some elements exist as gases but only two elements, mercury (Hg) and bromine (Br) are liquids at room temperature (approx 25oC). Hope this helped you!
7 0
3 years ago
45 Gg = [? ]x10!?g<br> ]<br> =<br> ]
nlexa [21]

1. The coefficient (green) is 45

2. The exponent (yellow) is 9

<h3>Conversion scale</h3>

To convert from giga grams (Gg) to grams (g), the following coversion scale can be use:

1 Gg = 10⁹ g

With the above convesion scale, we can convert 45 Gg to g as follow

1 Gg = 10⁹ g

Therefore,

45 Gg = 45×10⁹ g

Thus, 45 Gg is equivalent to 45×10⁹ g. Hence, we can conclude as follow:

  • The coefficient (green) is 45
  • The exponent (yellow) is 9

Learn more about conversion:

brainly.com/question/21919505

6 0
3 years ago
Suppose you have 100 grams of a radioisotope with a half-life of 100 years. How much of the isotope will you have after 200 year
Ulleksa [173]
        Amount Remaining     Years       #half lives 
             100g                            0                 0
             50 g                           100                1                  
             25g                            200               2                      




8 0
3 years ago
Read 2 more answers
Salts usually have a low melting point
Svetradugi [14.3K]
Yes they do if that was your question
8 0
3 years ago
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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
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