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vladimir2022 [97]
2 years ago
7

Which of the following metals reacts with oxygen and acids but is not very reactive with water?

Physics
1 answer:
Lina20 [59]2 years ago
7 0

Calcium is reactive with oxygen, acid, and water. Copper is not reactive with water as well as with acids, whereas magnesium is reactive with acid but reactive with water.

<h3>What is a Chemical Reaction?</h3>

One or more chemicals (the reactants) are changed into one or more other compounds during a chemical reaction (the products).Substances are made up of chemical components or chemical elements.

Each calcium atom donates a pair of electrons to an oxygen atom during the interaction between calcium and oxygen to create calcium oxide.

When calcium reacts with hydrochloric acid, it releases bubbles of hydrogen gas, and when large chunks of calcium react with hydrochloric acid, it releases hydrogen gas. Calcium in reaction with water produces calcium hydroxide and hydrogen gas

When magnesium and oxygen interact, they produce light that is so strong that it temporarily impairs your vision. Magnesium burns so brilliantly because it produces a lot of heat while doing so.

Magnesium reacts with hot water or water vapor to produce hydrogen gas and magnesium hydroxide.

Acid does not react with copper. Metals such as lead, copper, silver, and gold do not react with water at all.

To get more info about the chemical reaction:

brainly.com/question/3461108

#SPJ1

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A nucleus splits to form two or more smaller nuclei during which process?
salantis [7]

Answer:

nuclear fission

7 0
3 years ago
What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is 5.90 x 10^5 K?
OLEGan [10]

<u>Answer:</u> The average kinetic energy of helium atoms is 1.222\times 10^{-17}J

<u>Explanation:</u>

To calculate the average kinetic energy of the atom, we use the equation:

K=\frac{3}{2}kT

where,

K = average kinetic energy = ?

k = Boltzmann constant = 1.3807\times 10^{-23}J/K

T = temperature = 5.9\times 10^5K

Putting values in above equation, we get:

K=\frac{3}{2}\times 1.3807\times 10^{-23}J/K\times 5.9\times 10^5K\\\\K=1.222\times 10^{-17}J

Hence, the average kinetic energy of helium atoms is 1.222\times 10^{-17}J

4 0
3 years ago
A typical ten-pound car wheel has a moment of inertia of about 0.35kg⋅m2. The wheel rotates about the axle at a constant angular
QveST [7]

Answer:

The rotational kinetic energy of the rotating wheel is 529.09 J  

Explanation:

Given;

moment of inertia I = 0.35kg⋅m²

number of revolutions = 35.0

time of revolution, t = 4.00 s

Angular speed (in revolution per second), ω = 35/4 = 8.75 rev/s

Angular speed (in radian per second), ω = 8.75 rev/s x 2π = 54.985 rad/s

Rotational kinetic energy, K = ¹/₂Iω²

Rotational kinetic energy, K = ¹/₂ x 0.35 x (54.985)²

Rotational kinetic energy, K = 529.09 J  

Therefore, the rotational kinetic energy of the rotating wheel is 529.09 J  

7 0
3 years ago
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

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3 years ago
What is the difference between unicellular and multicellular organisms?
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