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3 years ago

Please help this is a test!! I’ll give 10 points

Chemistry

1 answer:

marshall27 [118]3 years ago

6 0

Answer:

I would say a person pulling out a garden hose.

Explanation:

The two forces in the first example is the boy lifting himself and gravity.

The two forces in the second example is the boys pushing and the sleds counterweights

The two forces in the fourth example is one dog pulling and the other pulling the other way

This makes it so the only possible answer is #3

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Help needed Assignment is due Justify that H2SO4 is Arrhenius acid and KOH is Arrhenius base.

irakobra [83]

Answer:

Arrhenius Acid:

According to Arrhenius concept, Acids are proton donors.

Since H₂SO₄ have a proton (H⁺ ion) and it can donate it to be made a sulphate ion, So it is an Arrhenius acid.

See the following reaction =>

H₂SO₄ + H₂O => HSO₄ + H₃O⁺

Arrhenius Base:

An Arrhenius base is a a proton acceptor.

KOH accepts the proton to to made to KOH₂ and a proton acceptor.

See the following reaction =>

KOH + H₂o => KOH₂ + OH⁻

6 0

3 years ago

Select the equations below that represent physical changes.

Ierofanga [76]

Physical changes- any change in the physical properties of the substance is a physical change. Generally physical properties include color, state, size , shape, odour, appearance . In any phyical change, no new substance is formed.

Chemical changes- any change in the chemical properties of the substance is a chemical change. Chemical properties changes only when something new is formed during a reaction.

∴ Part A- 2 H2O(l) → 2 H2(g) + O2(g)

In this, H2 and O2 is produced during the reaction which is different from H20. Thus, it is a chemical change.

Part B- H2O(l) → H2O(s)

In this, only the state of water changes from liquid to solid and no new product is formed. Thus, it is a physical change.

Part C-CO2(s) → CO2(g)

In this also, only the state of CO2 changes from solid to gas and no new product is formed. Thus, it is a physical change.

Part D-H2(g) → 2 H(g)

In this reaction, H2 molecule is dissociated into 2 hydrogen atom leading to formation of new products. Thus, it is a chemical change.

Finally, equations that represent physical changes are - B and C

B. H2O(l) → H2O(s)

C. CO2(s) → CO2(g)

6 0

4 years ago

summarize rutherford's model of an atom, and explain how he developed thus model based on the results of his famous gold-foil ex

Ainat [17]

Rutherford's atomic structure model was revolutionary. Contrary to J.J. Thompson's "plum pudding" model (which consisted of a solid, even mixture of protons and electrons), Rutherford's model consisted of one small, positively charged, dense nucleus, a layer of empty space, and a layer of negatively charged electrons. He came to this conclusion through his gold-foil experiment. He shot a ray of alpha particles towards the thin gold foil, and to Rutherford's surprise, some of the rays reflected back instead of going straight through the foil as he originally thought.

3 0

3 years ago

Space shuttle engines combine liquid hydrogen and liquid oxygen to produce an explosive reaction which releases great amounts of

Anastasy [175]

H2 + O -> H2O

I'm pretty sure that's the answer, just tell me if I'm wrong.

6 0

3 years ago

A rubber ball is held 4.0 m above a concave spherical mirror with a radius of curvature of 1.5 m. At time equals zero, the ball

exis [7]

The time elapsed when the ball placed above the concave mirror and the image formed would be at the same location is 0.55 s.

<h3>Image distance</h3>

The position of the image formed is determined using the followimg mirror formula;

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

where;

f is the focal length of the mirror
v is the image distance
u is the object distance

f = R/2

f = 1.5/2

f = 0.75 m

When the ball and its image is in the same position, u = v

The position of the ball is calculated as;

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u} \\\\\frac{1}{f} = \frac{2}{u} \\\\u = 2f\\\\u = 2(0.75)\\\\u = 1.5 \ m$

<h3>Time of motion of the ball</h3>

The time taken for the ball to travel the caluclated distance is determined as;

h = ut + ¹/₂gt²

1.5 = 0 + ¹/₂(9.8)t²

1.5 = 4.9t²

t² = 1.5/4.9

t² = 0.306

t = 0.55 s

Thus, the time elapsed when the ball and its image are at the same location is 0.55 s.

Learn more about concave mirror here: brainly.com/question/7512320

8 0

2 years ago