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zhuklara [117]
3 years ago
15

In a chemical equation, the chemicals that react are considered . In a chemical equation, the chemicals that are produced are co

nsidered . In chemical equations, both sides of the equation must be ; if they are not, coefficients must be used.
Physics
2 answers:
vovangra [49]3 years ago
4 0

Answer

Hi,

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation.

Good luck!

BartSMP [9]3 years ago
3 0
<h2>Answer: </h2>

<u>A chemical equation</u><u>  that shows the chemical formulas of substances that are reacting and the substances that are produced.</u>

<h2>Explanation:</h2>

A chemical equation should also be balanced on both sides. A chemical equation shows the substances involved in a chemical reaction . The substances that react are called reactants and the substances that are produced are called products.  A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.

For example

2 HCl + 2 Na → 2 NaCl + H 2

This equation will be counted  as two HCl plus two Na yields two NaCl and H two.

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A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A
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The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} is approximately  6,035.7 lb

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The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = T_{BC}

The tension in rope segment BD = T_{BD}

The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0...........................(2)

Which gives;

T_{BC} × sin(26.0°) = - T_{BD} × sin(21.0°)

T_{BC} = - T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ - T_{BD} × 0.8175

Substituting the value of, T_{BC}, in equation (1), gives;

- T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

- T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

T_{BD} ×0.1988 = 1200 lb

T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

T_{BD} ≈ 6,035.6938 lb

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T_{BC} ≈ -4934.1733 lb

From which we have;

The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

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