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3 years ago

Why feo is basic oxide

Chemistry

1 answer:

MAXImum [283]3 years ago

3 0

Answer:

A basic oxide is an oxide which when combined with water gives off a base. When a substance reacts chemically, both as a base or acid it termed as an amphoteric solution. Neutral Oxide is one which neither has an acidic characteristic or a basic one. Metal Oxides have an oxidation number of -2 and generally comprise of an oxygen anion.

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Did i answer these right??

WITCHER [35]

All of them are right except the neutron one. The neutrons is actually 176 because you subtract the number of protons from the atomic mass to find this.

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3 years ago

A certain shade of blue has a frequency of 7.00 × 1014 Hz. What is the energy of exactly one photon of this light?

NeTakaya

Energy = Planck's constant * Frequency

E = (6.62607004 × 10⁻³⁴<span>) * 7 * 10</span>¹⁴

E = 46.38 * 10⁻²⁰

E = 4.638 * 10⁻¹⁹ J

Hope this helps!

8 0

3 years ago

Read 2 more answers

what are the names of the "subunits" within the myofibrils that run the length of skeletal muscle fibers?

Shtirlitz [24]

Answer:

Sarcomeres

Explanation:

A muscle fiber is composed of many myofibrils, A sarcomere is the basic unit of a myofibril.

Sarcomeres are arranged in a stacked pattern throughout the muscle tissue, giving it a striped pattern. Sarcomeres are composed of thick and thin filaments. The thin and thick filaments interact initiate the movement of muscle.

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3 years ago

100 POINTS! PLEASE HELP! QUESTIONS ARE IN THE PICTURES. WILL GIVE BRAINLIEST, 5-STAR RATING, and THANKS. HALF-DONE ANSWERS WILL

Elenna [48]

Answer:

See Explanations....

Explanation:

NOTE => See attached graph.

1. 25-Turn Electromagnet

a. Strength of Batteries 1.5v vs 6.0v

6v:1.5v => 4:1 voltage ratio => That is, 6v battery is 4x voltage of 1.5v battery.

b. Ratio of Paperclips Picked Up

23clips(6v):6clips(1.5v) => 3.8clips(6v):1clip(1.5v)

That is, the 6 volt battery-EM picks up 3.8x more clips than the 1.5v battery-EM assuming constant number of coils given.

c. Empirical Relationship

From data given, an increase in voltage (V) produces a proportional increase in number of paperclips (C) picked up. Such is a Direct Relationahip between #Clips Picked Up and Battery Voltage.

Thus, C ∝ V => C = kV => k = C/V

Therefore, for k₁ = k₂ => C₁/V₁ = C₂/V₂

2. 50-Turn Electromagnet

a. Ratio of turns between electromagnets.

50-Turn EM : 25-Turn EM => 2:1 ratio of turns between electromagnets

b. Ratio of Paperclips Picked UP

Between maximum voltage and minimum voltages for 50-Turn EM

=> 47 clips (6v) : 13 clips (1.5v) => 3.6:1 ratio...

That is, the 6v battery-EM, on the average, picks up 3.6x more clips than the 1.5v battery-EM given 50-Turn EM.

3. Doubling number of coils:

Doubling the number of coils, on the average, doubles the number of clips picked up by either electromagnet.

Voltage 50-Turn EM 25-Turn EM Pick Up Ratio

1.5v 13 6 2:1

3.0v 26 12 2:1

4.5v 31 15.5 2:1

6.0v 47 23 2:1

4. Doubling Voltages:

Doubling voltage doubles the number of clips picked up by a given electromagnet regardless of the number of coils associated with magnet.

25-Turn EM Voltage Clips Pkd UP

1.5v 6

3.0v 12

Double voltage => Doubles # clips picked up*

50-Turn EM Voltage Clips Pkd UP

1.5v 13

3.0v 26

Double voltage => Doubles # clips picked up*

*Double Voltage => Double #clips picked up regardless of number of coil turns in electromagnet.

5. Predicting number of clips for 7.5v battery-EM

Using linear trendline equations shown on attached graphs...

#clips 50-Turn EM = 7.1333V + 2.5 = 7.1333(7.5) + 2.5 ≅ 56 clips

#clips 25-Turn EM = 3.6333V + 0.5 = 3.6333(7.5) + 0.5 ≅ 27 clips

Overall => 2:1 ratio of clips between 50-Turn EM and 25-Turn EM

6. Slope values based upon trendline equations shown on attached graphs are ...

25 Turn EM => slope = 3.6333

50 Turn EM => slope = 7.1333

Approximately a 2x increase upon doubling the number of coils of electromagnet.

Predicting number of clips for 1V electromagnets:

#Clips (1volt) 25-Turn EM = 3.6333(1) + 0.5 = 3.6833 ≅ 4 clips

#Clips (1volt) 50-Turn EM = 7.1333(1) + 2.5 = 9.6333 ≅ 10 clips

Shows to be consistent with double #clips with double number of coil turns.

6 0

3 years ago

In boyle's law, the initial volume of a gas is 120 ml at 256 kpa. what will be the volume if the pressure increases to 385 kpa?

Vinil7 [7]

<h3>Answer:</h3>

79.8 ml

<h3>Explanation:</h3>

Boyle's law describes the relationship between pressure and volume when the temperature is constant. For this question, I will be rounding to significant figures.

Boyle's Formula

When describing the relationship between pressure and volume, the formula $\displaystyle P_{1} V_{1}= P_{2} V_{2}$ . In this formula, $P_{1}V_{1}$ is the initial pressure and volume. On the other side, $P_{2} V_{2}$ is final pressure and volume. So, to find a missing variable you must plug in the values you are given.