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IRISSAK [1]
3 years ago
11

PLEASE HELP ASAP Which one of the following is a step used for balancing chemical equations?

Physics
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:

C. Count the atoms in each substance in the reactants and products.

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

In order for a chemical equation to be balanced, the condition which must be met is that the number of atoms in the reactants equals the number of atoms in the products.

This ultimately implies that, the mass and charge of the chemical equation are both balanced properly.

In Chemistry, all chemical equation must follow or be in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed by either a physical transformation or a chemical reaction but transformed from one form to another in an isolated (closed) system.

One of the step used for balancing chemical equations is to count the atoms in each substance in the reactants and products.

For example;

NH3 + O2 -----> NO + H2O

The number of atoms in each chemical element are;

For the reactant side:

Nitrogen, N = 1

Hydrogen, H = 3

Oxygen, O = 2

For the product side;

Nitrogen, N = 1

Hydrogen, H = 2

Oxygen, O = 2

When we balance the chemical equation, we would have;

NH3 + 3O2 -----> 4NO + 2H2O

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To make ice, a freezer that is a reverse Carnot engine extracts 37 kJ as heat at -17°C during each cycle, with coefficient of pe
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5 0
3 years ago
An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a
Zolol [24]

Answer:

a) \omega = 0.421\,\frac{rev}{s}, b) \Delta \theta = 0.066\,rev, c) v = 0.966\,\frac{mm}{s}, d) a = 3.293\,\frac{mm}{s^{2}}

Explanation:

a) The angular velocity of the turntable after 0.200\,s.

\omega = \omega_{o} + \alpha\cdot \Delta t

\omega = 0.240\,\frac{rev}{s}  + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)

\omega = 0.421\,\frac{rev}{s}

b) The change in angular position is:

\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot  \alpha \cdot t^{2}

\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}

\Delta \theta = 0.066\,rev

c) The tangential speed of a point on the rim of the turn-table:

v = r\cdot \omega

v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )

v = 9.655\times 10^{-4}\,\frac{m}{s}

v = 0.966\,\frac{mm}{s}

d) The tangential and normal components of the acceleration of the turn-table:

a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )

a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}

a_{t} = 2.078\,\frac{mm}{s}

a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}

a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

a_{n} = 2.554\,\frac{mm}{s^{2}}

The magnitude of the resultant acceleration is:

a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}

a = 3.293\,\frac{mm}{s^{2}}

8 0
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