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3 years ago

Which of these is a chemical change?

Physics

2 answers:

Radda [10]3 years ago

7 0

Anything that says burned is a chemical reaction or a chemical change. Here's a burning reaction.

C2H5OH + 302 ===> 2CO2 + 3H2O

This is the equation for burning alcohol (the kind you drink). When burned, the alcohol's reaction leaves a beautiful blue flame along with CO2 and H2O on the right side.

So B is the answer.

You're not doing anything to paper if you shred it except make it smaller. It retains all of its chemical properties and almost all its physical properties.

Ice melting doesn't do anything to the water when it changes phase except you now have a puddle on the table where the ice (cube) was. But you are still dealing with water.

Metal foil is heated, again the foil is left with all its chemical properties in tact. You can reuse the foil if you are cooking potatoes in an open fire. Nothing has altered the chemical properties of the foil.

Iteru [2.4K]3 years ago

6 0

The answer is when GAS IS BURNED because when gas is burned, it cannot return into its previous state. Also, new substances are formed.

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3 years ago

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Christina drives his moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. What distance did she cove

Mashutka [201]

Answer:

She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

Explanation:

Given that,

Christina drives his moped 7 kilometers North and stop for lunch and then drive 5 km east.

We need to calculate the total distance

Using formula of distance

$d=d_{1}+d_{2}$

Put the value into the formula

$d=7+5$

$d=12\ km$

We need to calculate the magnitude of displacement

Using formula of displacement

$D=xi+yj$

$D=5i+7j$

$D=\sqrt{5^2+7^2}$

$D= 8.6\ km$

The direction of her displacement is north east.

Hence, She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

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3 years ago

18. The orbits of planets being elliptical was one the planetary laws developed by

TEA [102]

It was a man named <span>Johannes Kepler. </span>

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3 years ago

Which formula represents final velocity of an object with average acceleration?

zheka24 [161]

Answer:

The equation v – = v 0 + v 2 v – = v 0 + v 2 is reflects the fact that when acceleration is constant, v – is just the simple average of the initial and final velocities.

Explanation:

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5 0

2 years ago

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Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago