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Reika [66]
3 years ago
13

Two examples of organic chemical

Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

there are literally millions - alkanes, alkenes, alkynes, alcohols, esters, carboxylic acids... to name but a few.

Explanation:

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The density of mercury is 13.6 g/cm'. How many quarts does 83.0 g of Hg occupy? (1 L=1.06 qt)​
forsale [732]

Answer:

0.0064 qt

Explanation:

Given data:

Density of mercury = 13.6 g/cm³

Mass of mercury = 83.0 g

How many quarts mercury have = ?

Solution:

First of all we will calculate the volume in cm³.

d = m/v

13.6 g/cm³ = 83.0 g / volume

Volume = 83.0 g /  13.6 g/cm³

Volume = 6.1 cm³

cm³ to L:

6.1 cm³ × 1 L / 1000  cm³

0.0061 L

In quarts:

1 L = 1.06 qt

0.0061 L × 1.06 qt / 1 L

0.0064 qt

4 0
3 years ago
HBrO3 <br> A. All non metals “acid subcategory”
Irina-Kira [14]
Non metals acid subcategories
4 0
3 years ago
What characteristic frequencies in the infrared spectrum of your estradiol product will you look for to determine whether the ca
gladu [14]

Answer:

Strong broad peak around 3200-3600 cm-1 should be present

Strong peak around around 1700 cm-1 should be absent

Explanation:

Infrared spectroscopy is an analytical technique which is used for molecular structure characterization by identifying the functional groups present in a given molecule based on the absorption wavelength (or wavenumber).

In an IR spectrum the carbonyl group is associated with the C=O stretch which occurs as a strong peak around around 1700 cm-1. For alcohol the -corresponding O-H stretching frequency occurs as a strong broad peak between 3200-3600 cm-1.

Therefore, in the case of estradiol the presence a strong broad peak in the 3200-3600 cm-1 and the absence of the peak at around 1700 cm-1. would suggest that the transformation is complete.

8 0
4 years ago
Consider the unbalanced equation for the oxidation of butene. C4H8 + 6O2 Right arrow. CO2 + H2O For each molecule of C4H8 that r
antiseptic1488 [7]
Answer is C4H8 + 6O2 —> 4CO2 + 4H2O
4 0
3 years ago
Iron-59 is used in medicine to diagnose blood circulation disorders. The half life of iron-59 is 44.5 days. How much of a 2.000
kolezko [41]

Answer:

0.258 mg of iron remains.

Explanation:

To solve this problem we can use the formula

M₂ = M₀ * 0.5^{\frac{t}{44.5} }

Where M₂ is the mass remaining, M₀ is the initial mass, and t is time in days.

Using the data given by the problem:

M₂ = 2.000 mg * 0.5^{\frac{133.5}{44.5} }

M₂ = 0.258 mg

7 0
3 years ago
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