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dezoksy [38]
3 years ago
14

I'm a given chemical reaction, the energy of the products is less than the energy of the reactants. Which statement is true for

this chemical reaction?
A energy is absorbed in the reaction
B energy is released in the reaction
C there is no transfer of energy in the reaction
D energy is lost in the reaction
Physics
2 answers:
garik1379 [7]3 years ago
8 0
The answer would have to be D.
Lady bird [3.3K]3 years ago
5 0

Answer: Option (B) is the correct answer.

Explanation:

A chemical reaction in which there is release of energy into the atmosphere is known as an exothermic reaction.

Also, energy of products is less than the energy of reactants for an exothermic reaction.

For example, A + B \rightarrow C + D + Heat

Whereas a chemical reaction in which heat is absorbed by the reactant molecules is known as an endothermic reaction.

Also, energy of products is more than the energy of reactants in an endothermic reaction.

For example, A + B + Heat \rightarrow C + D

Hence, we can conclude that the statement, energy is released in the reaction is true for this chemical reaction.

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6 0
3 years ago
The first five questions refer to the following problem:
Kruka [31]

Answer:

<h2>Δd=d2−d1</h2>

Explanation:

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If you lift weights faster than normal you have increased what?
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3 years ago
How many joules of work are done on a box when a force of 25 N pushes it 3 m?
Anika [276]

Answer:

75joules

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7 0
3 years ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va
WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V =  3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0
3 years ago
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