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3 years ago

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Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s) The unbalanced equation for the reduction of silver nitrate by copper is shown here. Wh

at is the coefficient for Cu(NO3)2 in the balanced equation?

1 answer:

cricket20 [7]3 years ago

3 0

Answer:

1

Explanation:

Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag

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Did you remember to put the lid back on the jar of mayonnaise?

Explanation: Hope this helps :)

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When iron rusts, iron atoms are destroyed true or false?

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3 years ago

ILL MARK BRAINLIEST <br>PLEASE HELP <br>2.Write the formula associating energy with power

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Answer:

The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters.

Explanation:

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3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

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The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

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3 years ago

(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i

ddd [48]

Answer:

a) $P = 370.993\,kPa$ , b) $F = 25.948\,kN$

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

$P = P_{atm} + P_{man}$

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$

$P = 370.993\,kPa$

b) The force exerted by the water is:

$F = (P - P_{atm})\cdot A$

$F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}$

$F = 25.948\,kN$

5 0

3 years ago

Read 2 more answers