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3 years ago

Temperature is a measure of

Chemistry

2 answers:

Triss [41]3 years ago

8 0

Temperature is a measure of "Molecular movement"

In short, Your Answer would be Option B

Hope this helps!

Cloud [144]3 years ago

4 0

I believe the answer is D, because heat is stored and used as a source of energy. I cant be C because temperature has nothing to do with chemicals and the other 2 are not important.

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PLEASE HELP! WILL MARK BRAINIEST IF ANSWERED CORRECTLY.

ioda

Answer:

H2O is a polar molecule

the H is slightly positive and the O is slightly negative so the Na is attracted to the O side of the water molecule and the Cl is attracted to the H side of the water molecule

Explanation:

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3 years ago

22.4 L is the volume of any gas regardless of atmospheric conditions.<br><br> O True<br><br> O False

Contact [7]

This is false. One mole of a gas occupies 22.4 L at STP, which is taken to be 0°C (273 K) and 1 atm. If atmospheric conditions depart from these values, this assumption cannot be used.

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3 years ago

URGENT

Vadim26 [7]

Answer:

Explanation:

Discussion

When Pressure increases equilibrium shifts to the side with the smallest number of moles. But which side is that?

N2(g) + 3H2(g) ⇌ 2NH3(g)

The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols

on the left side.

The right side has 2 mols of NH3 = 2 mols on the right.

Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.

Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.

2NH3 is only two mols.

The equilibrium shifts Right

Answer

7 0

2 years ago

One plus one hundred and thirty five

Makovka662 [10]

Answer:

Explanation:

136

thxx

5 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