3 years ago

Which sugar is the sweetest a) fructoseb) sucrosec) glucosed)lactose

Chemistry

2 answers:

qaws [65]3 years ago

7 0

Fructose is d sweetest sugar of all

Y_Kistochka [10]3 years ago

5 0

Fructose is the sweetest sugar of your options.

So the correct answer is A.

Hope this helps,

Davinia.

You might be interested in

How many moles of C6Cl6 are in 5.44g

Kazeer [188]

Explanation:

rbrhrhyhggggsdffffffffffv

4 0

3 years ago

The temperature in Jersey City, NJ in mid-Spring is 60*F. The temperature in Miami, FL is 90*F on the same day. Miami is about 1

timurjin [86]

Answer:

Temperature gradient = 30

90F - 60F = 30F

The temperature gradient is 30F.

If I am right, let me know.

7 0

3 years ago

Read 2 more answers

is this information about magnesium true symbol..mg number..12 atomic mass..24.305 number of protons 12.... PLZ HELP :)

maxonik [38]

Yes, that is all correct!

The atomic number is the same as the number of protons (12).
Magnesium = Mg
Atomic mass is 24. 305.

3 0

3 years ago

What major events cause islands to form in the pacific ocean

zmey [24]

One of the major events is Volcanos etc...

4 0

3 years ago

Read 2 more answers

What is the boiling point of a solution produced by adding 610 g of cane sugar (molar mass 342.3 g/mol) to 1.4 kg of water? For

wel

Answer:

Boiling point of solution is $100.65^{0}\textrm{C}$

Explanation:

Cane sugar is a non-volatile solute.

According to Raoult's law for a non-volatile solute dissolved in a solution-

$\Delta T_{b}=K_{b}.m$

Where, $\Delta T_{b}$ is elivation in boiling point of solution, $K_{b}$ is ebbulioscopic constant of solvent (how much temperature is raised for dissolution of 1 mol of non-volatile solute) and m is molality of solution.

Here, $K_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}$

610 g of cane sugar = $\frac{610}{342.3}$ moles of cane sugar

= 1.78 moles of cane sugar

So, molality of solution (m) = $\frac{1.78}{1.4}mol.kg^{-1}=1.27mol.kg^{-1}$

Plug in all the values in the above equation, we get-

$\Delta T_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}\times 1.27mol.kg^{-1}=0.65^{0}\textrm{C}$

So, boiling point of solution = $(100+0.65)^{0}\textrm{C}=100.65^{0}\textrm{C}$

7 0

3 years ago