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3 years ago

7

Cellular respiration is an example of a chemical reaction where energy is released. Which of the following is a reaction where e

nergy is released?
A-Exothermic reaction

B-Endothermic reaction

C-Dissolution reaction

D-Acid-Base reaction

1 answer:

SSSSS [86.1K]3 years ago

6 0

C. Dissolution reaction

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A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average

Maksim231197 [3]

Answer: $72.66 AU=1.089(10)^{10} km$

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period $T$ of a planet is proportional to the cube of the semi-major axis $a$ of its orbit”:

$T^{2}\propto a^{3}$ (1)

Now, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes:

$T^{2}=a^{3}$ (2)

So, knowing $T=619.36 years$ and isolating $a$ from (2) we have:

$a=\sqrt[3]{T^{2}}$ (3)

$a=\sqrt[3]{(619.36 years)^{2}}$ (4)

Finally:

$a=72.66 AU$ T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

$a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km$

4 0

3 years ago

Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h

Kaylis [27]

distance to the star Betelgeuse: 640 ly

As we know that

$1 ly = 63000 AU$

also we know that

$1AU = 1.5 \times 10^8 km$

$1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km$

So the distance of Betelgeuse = 640 ly

$d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m$

distance to the star VY Canis Majoris: $3.09 × 10^8 AU$

$d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km$

$d_2 = 4.64 \times 10^{16} km$

distance to the galaxy Large Magellanic Cloud: 49976 pc

$1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km$

$1pc = 3.08 \times 10^{13} km$

now we have

$d_3 = 49976 \times 3.08 \times 10^{13}$

$d_3 = 1.54 \times 10^{18} km$

distance to Neptune at the farthest: 4.7 billion km

$d_4 = 4.7 \times 10^9 km$

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

6 0

3 years ago

What is the anion of the ionic compound NaC2H3O2<br><br> thank you!

KIM [24]

C₂H₃O₂⁻ is an anion.

<u>Explanation:</u>

NaC₂H₃O₂(s) → Na⁺(aq) + C₂H₃O₂⁻(aq)

NaC₂H₃O₂ when dissociated, yields Na⁺ and C₂H₃O₂⁻.

Anion is a negatively charged ion.

In this case, C₂H₃O₂⁻ is an anion.

6 0

3 years ago

What happens to the particles of a liquid when energy is removed from them?

KonstantinChe [14]

Answer:

D: The distance between the particles decreases

Explanation:

Taking away energy slows down molecules, like how you slow down when you are cold (I think)

3 0

3 years ago

The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$

$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

$A = 4.3 \times 10^{-3} m^2$

$v = 40.2 m/s$

so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

$F = 1.46 N$

So acceleration of the ball is

$a = \frac{F}{m}$

$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 40.2^2 = 2(-10.1)(18.4)$

$v_f = 35.3 m/s$

4 0

3 years ago