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1. What coefficients would balance the following equation?

maksim [4K]

Answer:

D. 4Al + 3O2 → 2Al2O3

Explanation:

Chemical reactions involves the chemical combination of two or more substances called REACTANTS to yield other substances called PRODUCTS. However, in accordance with the LAW OF CONSERVATION OF MASS, the amount of reactants must be equal to that of the products.

To accomplish this, the reaction must be BALANCED. A balanced equation is an equation in which the number of atoms of each element in the reactant side equals the number of atoms in the product side. In this reaction involving Aluminum and Oxygen to give Aluminum oxide as follows:

Al + 02 → Al2O3

A coefficient is used to balance the number of atoms on both sides of the equation as follows:

4Al + 3O2 → 2Al2O3

3 0

3 years ago

Read 2 more answers

What is newton first law

andrew11 [14]

Answer:

Newtons first law states that:

If a body is in rest or motion in a straight line, it remains at rest or at motion in a straight line with constant speed until and unless and external unbalanced force acts on it.

'This law is also known as the law of Inertia.'

5 0

2 years ago

A particular lightbulb is designed to consume 40 W when operating on a car's 12-V DC electric power. If you supply that bulb wit

Andrew [12]

40V because it will provide the same amount of power.

4 0

2 years ago

Any 3 differences between telescope and microscope

Keith_Richards [23]

An instrument used to observe or imagine very small object using an optical mangifier
mirco cell.
Telescope is a magnifer of distance object

4 0

3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: