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LenaWriter [7]
2 years ago
14

What is the law of conservation of energy as it applies to exothermic dissolution processes?Energy given off by the system durin

g dissolution equals the energy absorbed by the surroundings.Energy given off by the system during dissolution is less than the energy absorbed by the surroundings.Energy given off by the system during dissolution is greater than the energy absorbed by the surroundings.Energy given off by the system during dissolution may be greater or less than the energy absorbed by the surroundings.
Chemistry
1 answer:
Pani-rosa [81]2 years ago
4 0

Answer:

The energy given off by the system during dissolution is less than the energy absorbed by the surroundings.

Explanation:

Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.

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A person throws a ball up into the air, and the ball falls back toward Earth. At which point would the kinetic energy be the low
dybincka [34]
It would be d bc when it’s at its lowest point it means not that much energy and that would be its lowest point which is D
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2 years ago
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Which part of experimental design is most important to a scientist when replicating an experiment? Having exactly the same data
yarga [219]

Answer:

The answer to your question is below:

Explanation:

Having exactly the same data as the previous experiment I think that having the same data as the previous experiment is extremely important but not the most important, for me is the second most important.

Using the same procedure and variables as the previous experiment For me, this is the most importan thing when a scientist is designing an experiment, because if he or she follow exactly the same procedure and variables, then the results will be very close.

Conducting an experiment similar to the previous experiment  This characteristic is important but not the most important.

Using the same laboratory that was used in the previous experiment It is not important the laboratory, if the procedure and variables are the same, your experiment must give the same results in whatever laboratory.

5 0
2 years ago
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An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the or
galben [10]
The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
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v = 5.367×10¹¹ m/s
4 0
2 years ago
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What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?
expeople1 [14]

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

8 0
1 year ago
Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bee
lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a)        130 g of C7H14O2 ---------------- 1 mol of C7H14O2

           1 x 10 ⁻⁶ g              ---------------      x

           x = 7.7 x 10 ⁻⁹ mol

          1 mol of C7H14O2   --------------   6 .023 x 10 ²³ molecules

          7.7 x 10⁻⁹ mol          --------------    x

          x = 4.64 x 10¹⁵ molecules

b)      130 g of C7H14O2   ----------------   1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2   -----------------     x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2    ---------------   2 mol of O2

        7.7 x 10 ⁻⁹                 ----------------   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2 -----------------  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸  ----------------   x

        x = 9.28 x 10 ¹⁵ atoms of O2

8 0
2 years ago
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