: Modeling Chemical Reactions

Physics

1 answer:

77julia77 [94]3 years ago

8 0

Answer:

A Matter is not created or destroyed and the same amount of mass remains

Explanation:

In a chemical reaction, according to the law of conservation of matter, "matter is neither created nor destroyed and the same amount of mass remains in such a process".

New kinds of products are formed in a chemical reaction because they undergo chemical changes.
The chemical changes results in energy changes and formation of new kinds of substances.
Based on the law of conservation of matter, mass is constant.
Mass is not lost or gained.

You might be interested in

By measuring the amounts of parent isotopes and daughter product in the minerals contained in a rock, and by knowing the half-li

sertanlavr [38]

The asker of the second question needs a tutorial in radiometric dating. There is little likelihood that the daughter isotope has the same atomic weight as the parent isotope. To measure the mass isotopes doesn't tell us how many atoms of each exist. To get around that let's pretend — which will likely serve the purpose ineptly intended — that the values give an the particle ratio, 125:875.

The original parent isotope count was 125 + 875 = 1000. The remaining parent isotope is 125/1000 or 1/8. 1/8 = (1/2)^h, where h is the number of half-lives. 

h = log (1/8) ÷ log(1/2) = 3 

And 3 half-lives • 150,000 years/half-life = 450,000 years.

4 0

3 years ago

PLEASE ANSWER ASAP!! A photographer in a helicopter ascending vertically at a constant rate of 9.5 m/s accidentally drops a came

vodomira [7]

C should be the answer! :)))

3 0

3 years ago

Read 2 more answers

A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k

Bad White [126]

Answer:

$v_{f,w} = 1.791\,\frac{m}{s}$ , $v_{f,c} = 0.972\,\frac{m}{s}$

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

$I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}$

$\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)$