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2 years ago

Which result occurs during an exothermic reaction?

1 answer:

3 0

In exothermic reactions, heat and light are released to the surrounding environment. On the other hand, in an endothermic reaction, heat is required and therefore it can be considered as a reactant.

In exothermic reactions, light and heat are released into the environment (Option D).

Exothermic reactions release energy in the form of heat or light.

Combustion reactions are generally exothermic reactions.

After an exothermic reaction takes place it is possible to observe that the energy of the products of the reaction is lesser than the energy of the reactants.

The energy released in exothermic reactions is evidenced by the increase in temperature of the reaction.

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A wind-tunnel experimentis performed on a 1/25scale model of a supersonic aircraft. The prototype aircraft flies at 450 m/s in c

KengaRu [80]

Answer: V = 504m/a

F = 4N

Explanation: please find the attached file for the solution

4 0

3 years ago

Q 1 . How many significant figures are in the following measurement? 0.0009(1 point)

Crazy boy [7]

Here we have some questions about experimental errors.

Q1) We want to see how many significant figures have the measure:

0.0009

The number of significant figures is the number of known digits that are not the leading zeros.

Here we can see four leading zeros, and a single-digit different than zero, which is a 9.

Then we have only one significant figure, the 9.

Q2) Here we will use the measure that is the less exact, as the error of that measure may be larger than the smaller significant figures of the other measures.

Then:

31.2 lb + 38.02lb + 45 lb

The worst measure is 45lb, so the smallest significant figure that we should use is the first one at the left of the decimal point, then we need to round the other two measures to the next whole number, we will get:

31 lb + 38 lb + 45 lb = 114lbs

Q3) We know that the measure is 11.5 seconds and the uncertainty of 1.7%, then the uncertainty will be the 1.7% of the above measure:

(1.7%/100%)*11.5 s = 0.1955 s

Notice that our measure has one significant figure after the decimal point, so we need to round the error to the same significant figure.

0.1955 s ≈ 0.2s

Then the measure is:

11.5 s ± 0.20 s

Q4) We have the measure:

312.0 mph ± 3.9 mph.

The percent uncertainty will be the quotient between the error and the measure times 100%, or:

(3.9 mph/312.0 mph)*100% = 1.25%

This is a percent error, we do not need to round this.

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5 0

3 years ago

Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both fil

MatroZZZ [7]

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴ ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴ ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

6 0

3 years ago

Compute the density in g/cm^3 of a piece of metal that had mass of 0.485 kg and a volume of 52cm^3

steposvetlana [31]

Answer:

9.3 g/cm³

Explanation:

First, convert kg to g:

0.485 kg × (1000 g / kg) = 485 g

Density is mass divided by volume:

D = (485 g) / (52 cm³)

D = 9.33 g/cm³

Rounding to two significant figures, the density is 9.3 g/cm³.

8 0

3 years ago

A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o

dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?

we know $p = \frac{v^2}{R}$ .....................(i)

we need to find change in power

$\Delta P = \frac{\Delta (V^2)}{R}$

$\Delta P = \frac{2 V \Delta V}{R}$ ..............(ii)

from equations we get

$\frac{\Delta P}{P} = \frac{2 \Delta V}{V}$

$\frac{\Delta P}{P} = 2 \frac{110 -120}{120}$

$\frac{\Delta P}{P} = -2(\frac{10}{120})$

$\frac{\Delta P}{P} = - 0.16 %$

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller

7 0

3 years ago