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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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Momentum is what combination of variables?
BigorU [14]

p=mv

where p=momentum, m=mass and v=velocity.

Hence momentum is the combination of mass and velocity.

Hope this helps you!

8 0
4 years ago
How many seconds did it take (after starting his descent) for the worker to hit the ground? Answer in units of s.
elena-14-01-66 [18.8K]

Answer:

This question is incomplete

Explanation:

The question is incomplete. However, to determine the time (in seconds) it took a worker to hit the ground from an elevated point. The speed the worker was coming with to the ground and the distance between the elevated point and the ground will have to be considered. Thus the formula to be used here will be

Speed (in meter per second) = distance (in meters) ÷ time (in seconds)

time (in seconds) = distance (in meters) ÷ speed (in meter per seconds)

6 0
3 years ago
A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does
Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right) (1)

Where:

y(t) - Position of the mass as a function of time, measured in meters.

A - Amplitude, measured in meters.

k - Spring constant, measured in newtons per meter.

m - Mass of the block, measured in kilograms.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

k = \frac{m\cdot g}{\Delta y} (2)

Where:

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

\Delta y - Deformation of the spring due to gravity, measured in meters.

If we know that m=1.65\,kg, g = 9.807\,\frac{m}{s^{2}} and \Delta y = 0.260\,m, then the spring constant is:

k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}

k = 62.237\,\frac{N}{m}

If we know that A = 0.130\,m, k = 62.237\,\frac{N}{m}, m=1.65\,kg, x(t) = 0\,m and \phi = 0\,rad, then (1) is reduced into this form:

0.130\cdot \cos (6.142\cdot t)=0 (1)

And now we solve for t. Given that cosine is a periodic function, we are only interested in the least value of t such that mass reaches equilibrium position. Then:

\cos (6.142\cdot t) = 0

6.142\cdot t = \cos^{-1} 0

t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s

t \approx 0.255\,s

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0
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