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I am Lyosha [343]
3 years ago
6

Check all that apply to sugar.

Chemistry
2 answers:
Sav [38]3 years ago
6 0

Answer:

Does not conduct electricity and is a non-electrolyte

Explanation:

Because I pay attention

harina [27]3 years ago
3 0

Answer:

Candy

Explanation:

its sweet and has a lot of sugar and acid

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The molarity of a solution prepared by dissolving 4.11 g of NaI in enough water to prepare 312 mL of solution is
Dimas [21]

Answer:

The correct answer is 8.79 × 10⁻² M.

Explanation:

Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,

No. of moles of NaI = Weight of NaI/ Molecular mass

= 4.11 / 149.89

= 0.027420

The vol. of the solution given is 312 ml or 0.312 L

The molarity can be determined by using the formula,

Molarity = No. of moles/ Volume of the solution in L

= 0.027420/0.312

= 0.0879 M or 8.79 × 10⁻² M

6 0
3 years ago
Khjgjhgfmhgxfhkugfdnhmvngchnvnhvhgh
igor_vitrenko [27]

Answer:

I don’t know

Explanation:

5 0
3 years ago
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Liquid helium boils at –268.93 °C. What is the boiling point of helium on the Kelvin temperature scale?
BARSIC [14]

In order to change celcius to kelvin always add 73 to it leaving you with -195.93

4 0
3 years ago
Read 2 more answers
Given the following equation: 2 K3N ---> 6 K + N2 How many moles of N2 can be produced by letting 12.00 moles of K3N react?
myrzilka [38]
The answer will be 12 nitrogen or N2 will be produced because if you changed the coefficient to 12 on the reactant side and distribute, the nitrogen would be 12 when distributed and what happens on one side has to equal to the other side, which is the product side.
8 0
3 years ago
Fatty acids spread spontaneously on water to form a monomolecular film. A solution containing 0.10 mm^3 of a fatty acid is dropp
lara31 [8.8K]

Divide the volume by the area. Using scientific makes things a bit cleaner.

0.10\,\mathrm{mm}^3 = 10^{-1}\,\mathrm{mm}^3

400.\,\mathrm{cm}^2 = 4\times10^2\,\mathrm{cm}^2

Then

\dfrac{10^{-1} \,\mathrm{mm}^3}{4\times10^2\,\mathrm{cm}^2} \cdot \dfrac{\left(\frac{1\,\rm m}{10^3\,\rm mm}\right)^3}{\left(\frac{1\,\rm m}{10^2\,\rm cm}\right)^2} = \dfrac{10^{-1}\times10^{-9} \,\mathrm m^3}{4\times10^2\times10^{-4}\,\mathrm m^2} = \dfrac{10^{-10}}{4\times10^{-2}}\,\mathrm m \\\\ ~~~~~~~~= 0.25\times10^{-8}\,\mathrm m

Now, 1 m = 10⁹ nm, so

0.25 \times10^{-8}\,\mathrm m \cdot \dfrac{10^9\,\rm nm}{1\,\rm m} = 0.25\times10^1\,\mathrm{nm} = \boxed{2.5\,\rm nm}

8 0
1 year ago
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