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schepotkina [342]
2 years ago
5

What is potential energy?

Chemistry
2 answers:
luda_lava [24]2 years ago
5 0

Answer:

B. The energy of position or composition

Explanation:

Potential energy is the energy possessed by an object or a body due to it's position in relation with other objects or bodies. It is a stored energy.

Vlad [161]2 years ago
4 0

Answer:

B

Explanation:

it is the energy stored in the chemical bonds of a substance

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Please help I don’t know if my answer is right!!
Temka [501]

Answer:

neutralization reaction! Aka: option C!

HOPE THIS HELPS! :)

Explanation:

8 0
3 years ago
8.43g of CaCl2·2H2O is dissolved in 40.0g of H2O. What is the molality of the solution? (Note that the H2O in the CaCl2·2H2O mus
lara31 [8.8K]

Answer:

[CaCl₂·2H₂O] = 1.43 m

Explanation:

Molality is mol of solute / kg of solvent.

Mass of solvent = 40 g

Let's convert g to kg → 40 g / 1000 = 0.04 kg

Let's determine the moles of solute (mass / molar mass)

8.43 g / 146.98 g/mol = 0.057 mol

Molality = 0.057 mol / 0.04 kg → 1.43

3 0
4 years ago
Read 2 more answers
Weather stations send up weather balloons with _____ to transmit temperature, humidity, wind speed, and wind direction measureme
laiz [17]
Technology? GPS? Analyzers? try those
6 0
3 years ago
Determine the mass in grams of 4.69 x 1021 atoms of barium. (The
Virty [35]

Answer:

1.07 g Ba

Explanation:

Hello there!

In this case, according to the definition of the Avogadro's number and the molar mass, it is possible to say that 6.022x10^{23} atoms of barium equal one mole, and at the same time, 1 mole equals 137.327 grams of this element; thus, it is possible to say that 6.022x10^{23} atoms of barium have a mass of 137.327 grams; therefore, it i possible for us to calculate the required mass in grams as shown below:

4.69x10^{21}atoms*\frac{137.327gBa}{6.022x10^{23} atoms} \\\\=1.07gBa

Best regards!

5 0
3 years ago
Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)


Rate Law: k[HI]^2


Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

5 0
4 years ago
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