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Anestetic [448]
2 years ago
5

Si2C19 What is this compound

Chemistry
1 answer:
lutik1710 [3]2 years ago
7 0

Answer:

BROMOTRIPHENYLSILANE

Explanation:

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What does CoH12O6 tell you about the glucose molecule?
Leokris [45]

Answer:

Glucose = C6H12O6

molecular mass = 6(12) + 12(1) + 6(16)

= 72 + 12 + 96

= 180 g

Explanation:

Glucose has a chemical formula of: C6H12O6 That means glucose is made of 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms. ... Glucose is produced during photosynthesis and acts as the fuel for many organisms.

8 0
2 years ago
<img src="https://tex.z-dn.net/?f=H_2PO_4%5E-%28aq%29%20%5Crightarrow%20H%5E%2B%28aq%29%20%2B%20HPO_4%5E%7B2-%7D%28aq%29" id="Te
klasskru [66]

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

4 0
3 years ago
How many grams are in 3.14 x 1015 molecules of CO?
Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15        Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15             Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x                        Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0
2 years ago
What are 4 examples of sediment that may become sedimentary rock
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7 0
3 years ago
5. The density of ethanol is 0.789 g/mL. Find the mass of a sample of ethanol that has a volume of 150.0 mL.
GenaCL600 [577]
. 00526 g is mass of ethanol
6 0
3 years ago
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