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fomenos
2 years ago
11

How many molecules of glucose are in 5. 72 grams of glucose c6h12o6?.

Chemistry
1 answer:
liq [111]2 years ago
4 0

Answer:

5.72 g ( 1 mol / 180.16 g ) (  6.022 x 10^23 molecules / mole ) = 1.90x10^23 molecules

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Slove this fill in blanks​
bagirrra123 [75]

Answer:

1) Heterogeneous mixture

2). homogeneous mixture

3) water is solvent and sugar is solute

4). sublimation process

3 0
3 years ago
Read 2 more answers
What is the molarity of a solution that has 2.50 moles of NaOH dissolved in 0.500 L of solution?
const2013 [10]

Answer: 5 is the molarity

Explanation:

The molarity formula is moles over liters and that in your case is 2.50 moles divided by .500 L which results in 5 which is your answear hope this helped god bless

5 0
4 years ago
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water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water
Gemiola [76]

Answer is: glycerol because it is more viscous and has a larger molar mass.

Viscosity depends on intermolecular interactions.

The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0
4 years ago
Which Carbon is the triple bound attached to in 6-ethyl-2-octyne?
soldi70 [24.7K]

Answer:

-second

Explanation:

6-ethyl-2-octyne is an unsaturated compound with a triple bond.

6-ethyl-2-octyne will have a triple bound attached to the second carbon. The suffix -yne suggests that compound carry a triple bond and the number  "2" before suffix refers to the position of triple bond that is second carbon.

Hence, the correct option is  "-second ".

7 0
3 years ago
Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.8 min. (A positr
Lera25 [3.4K]

Answer:

k = 6.31 x 10⁻³ min⁻¹

Explanation:

The equation required to solve this question is:

k = 0693 / t half-life

This equation is derived from the the equation from the radioctive first order reactions:

ln At/A₀ = -kt

where At is the number of isoopes after a time t , and A₀ is the number of of isotopes initially. The half-life is when the number of  isotopes has decayed by a half, so

ln(1/2) = -kt half-life

-0.693 = - k t half-life

t half-life = 109.8 min

⇒ k = 0.693 / t half-life = 0.693 / 109.8 min = 6.31 x 10⁻³ min⁻¹

6 0
3 years ago
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