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3 years ago

Write a hypothesis for this: Eating Breakfast increases performance in school

Chemistry

2 answers:

marishachu [46]3 years ago

8 0

If he eats breakfast, then he will have an increased school performance.

Gelneren [198K]3 years ago

7 0

Answer:

If someone were to eat breakfast before going to school, they would have an increased work performance while in school.

Explanation:

P.S. Brainliest is cool, you should totally hand one over.

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A characteristic feature of any form of chromatography is the ________.a. calculation of an Rf value for the molecules separated

podryga [215]

Answer : The correct option is, (c) use of a mobile and a stationary phase.

Explanation :

Chromatography : It is a separation process or technique of a mixture in which a mixture is distributed between the two phases at different rates, one of which is stationary phase and another is mobile phase.

Mobile phase : The mixture is dissolved in a solution is known as mobile phase.

Stationary phase : It is an adsorbent medium and It is a solid, liquid or gel that remains immovable when a liquid or a gas moves over the surface of adsorbent. It remains stationary.

Hence, a characteristic feature of any form of chromatography is the use of a mobile and a stationary phase.

4 0

3 years ago

liquid octane(CH3)(CH2)6CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water(H2O). I

max2010maxim [7]

The percent yield of carbon dioxide will be 49.0 %.

<h3>Percent yield</h3>

First, let's look at the equation of the reaction:

$2C_8H_1_8 + 25O_2 -- > 16CO_2 + 18H_2O$

The mole ratio of octane to oxygen is 2:25.

Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol

Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol

Thus, octane is limiting.

Mole ratio of octane to carbon dioxide = 2:16.

Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol

Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams

Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %

More on percent yield can be found here: brainly.com/question/17042787

#SPJ1

6 0

1 year ago

Why was Rutherford's Gold Foil Experiment so important?

Oksanka [162]

Answer:

Answer is 'B' I think

Explanation:

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.

4 0

3 years ago

three molecules of oxygen react with four molecules of hydrogen to produce water molecules write a balanced chemical equation

mixer [17]

Answer:

ExpC

→

This is the balanced reaction equation for the combustion of methane.

Explanation:

The Law of Conservation of Mass basically states that matter can neither be created nor destroyed. As such, we must be able to show this in our chemical reaction equations.

If you look at the equation above, you'll see an arrow that separates the reaction equation into two parts. This represents the direction of the reaction.

To the left of the arrow, we have our reactants.

To the right of the arrow, we have our products.

The quantity of each individual element in the left must equal the quantity of each individual element in the right.

So if you look below, you'll see the unbalanced equation, and I'll try to explain how to balance the reaction.

→

Our reactants in this equation are

and

Our next step is to break these down into individual atoms.

We have:

1 C atom, 4 H atoms & 2 O atoms.

If you're confused by this, look to see the little number to the bottom right of each element, the subscript, and it tells you how many of each atom are in the molecule. Make sense?

Now we look to the other side of the equation.

Here we see our products are

Again, we break these down into individual atoms again.

We have:

1 C atom, 2 H atom, 3 O atom

6 0

2 years ago

HELLPPP PLZZ ASAPP!

Sphinxa [80]

Answer:

Condensation and Depositon

Explanation:

Condensation is from gas to liquid

Deposition is from gas to solid

7 0

2 years ago