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podryga [215]
2 years ago
6

Draw what a chromosome looks like during metaphase. identify the chromatids and the centromere

Chemistry
1 answer:
netineya [11]2 years ago
8 0
This is what a chromosome looks like during mataphase

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Describe the three mechanisms of heat flow (conduction, convection, and radiation), and the factors that influence their rates.
trapecia [35]

Answer:

Just as interesting as the effects of heat transfer on a system are the methods by which it occurs. Whenever there is a temperature difference, heat transfer occurs. It may occur rapidly, as through a cooking pan, or slowly, as through the walls of a picnic ice chest. So many processes involve heat transfer that it is hard to imagine a situation where no heat transfer occurs. Yet every heat transfer takes place by only three methods:

Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on a macroscopic scale—we know that thermal motion of the atoms and molecules occurs at any temperature above absolute zero.) Heat transferred from the burner of a stove through the bottom of a pan to food in the pan is transferred by conduction.

Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in a forced-air furnace and in weather systems, for example.

Heat transfer by radiation occurs when microwaves, infrared radiation, visible light, or another form of electromagnetic radiation is emitted or absorbed. An obvious example is the warming of Earth by the Sun. A less obvious example is thermal radiation from the human body.

:)

Further more info...

Conduction:

As  you walk barefoot across the living room carpet in a cold house and then step onto the kitchen tile floor, your feet feel colder on the tile. This result is intriguing, since the carpet and tile floor are both at the same temperature. The different sensation is explained by the different rates of heat transfer: The heat loss is faster for skin in contact with the tiles than with the carpet, so the sensation of cold is more intense.

Convection :

In convection, thermal energy is carried by the large-scale flow of matter. It can be divided into two types. In forced convection, the flow is driven by fans, pumps, and the like. A simple example is a fan that blows air past you in hot surroundings and cools you by replacing the air heated by your body with cooler air. A more complicated example is the cooling system of a typical car, in which a pump moves coolant through the radiator and engine to cool the engine and a fan blows air to cool the radiator.

Radiation :

You can feel the heat transfer from the Sun. The space between Earth and the Sun is largely empty, so the Sun warms us without any possibility of heat transfer by convection or conduction. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. In these examples, heat is transferred by radiation. That is, the hot body emits electromagnetic waves that are absorbed by the skin. No medium is required for electromagnetic waves to propagate. Different names are used for electromagnetic waves of different wavelengths: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

7 0
3 years ago
Consider a circuit in which the left half-cell was prepared by dipping a Pt wire in a beaker containing an equimolar mixture of
Step2247 [10]

Answer:

a) Ti⁻²|Ti || Pt|Pt⁻²

b) +2.83 V

c) Pt + Ti⁻² → Pt⁻² + Ti

d) Pt

Explanation:

a) In the cell circuit, a redox reaction will happen, and one of the electrodes will oxide (it will lose electrons) and the other will reduce (it will gain electrons). The potential to reduce is measured by the standard potential reduction (E°). As higher is it, as easy is to the reduction reaction happens. The values for Pt and Ti are:

Pt + 2e⁻ → Pt⁻² E° = +1.20 V

Ti + 2e⁻ → Ti⁻² E° = -1.63 V

So, Pt will reduce and Ti will oxide. The oxidation reaction is the opposite of the reduction, and at the line notation, first is placed the oxidation reaction, thus:

Ti⁻²|Ti || Pt|Pt⁻²

b) The cell voltage (ΔE°) is the E° of the reduction reaction less the E° of the oxidation reaction:

ΔE° = 1.20 - (-1.63)

ΔE° = +2.83 V

c) The spontaneous net cell reaction will be the sum of the reduction reaction and the oxidation:

Pt + 2e⁻ → Pt⁻² (reduction)

Ti⁻² → Ti + 2e⁻ (oxidation)

----------------------------

Pt + Ti⁻² → Pt⁻² + Ti

d) In the cell, the electrodes are called cathode and anode. The electrode where the oxidation occurs is the anode, and the other, where the reduction occurs, is the cathode. Thus, in this case, the Pt electrode is the anode.

4 0
2 years ago
Describe the composition of Earth’s atmosphere
ahrayia [7]
Nitrogen accounts for 78% of the atmosphere, oxygen 21% and argon 0.9%. Gases like carbon dioxide, nitrous oxides, methane, and ozone are trace gases that account for about a tenth of one percent of the atmosphere
7 0
3 years ago
Wine goes bad soon after opening because the ethanol CH3CH2OH in it reacts with oxygen gas O2 from the air to form water H2O and
lyudmila [28]

Answer: 2.73g of CH3CH2OH Will be consumed

Explanation:

CH3CH2OH + O2 —> CH3COOH + H2O

MM of CH3CH2OH = 12 + 3 +12 + 2 16 +1 = 46g/mol

MM of O2 = 16 x2 = 32

Mass conc. Of O2 = 1.9g

From the equation,

32g of O2 consumed 46g of CH3CH2OH .

Therefore, 1.9g of O2 will consume Xg of CH3CH2OH i.e

Xg of CH3CH2OH = (1.9 x 46)/32 = 2.73g

3 0
2 years ago
A 29.05 gram sample of cobalt is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 40.88 g. Determ
deff fn [24]

The empirical formula of the oxide is Co₂O₃.

<em>Step 1</em>. Calculate the <em>mass of oxygen</em>

Your reaction is

 Cobalt + oxygen ⟶ cobalt oxide

29.05 g +    x g    ⟶     40.88 g

According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products. Thus,

29.05 g + <em>x</em> g ⟶ 40.88 g

<em>x</em> = 40.88 – 29.05 = 11.83

<em>Step 2</em>. Calculate the <em>moles of each element</em>

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Co to O.

<em>Moles of Co</em> = 29.05 g Co × (1 mol Co /(58.93 g Co) = 0.492 96 mol Co

<em>Moles of </em>O = 11.83 g O × (1 mol O/16.00 g O) = 0.739 38 mol O

<em>Step 3</em>. Calculate the <em>molar ratio</em> of the elements

Divide each number by the smaller number of moles

Co:O = 0.429 26:0.739 38 = 1:1.4999

<em>Step 4</em>. Multiply each number by a factor that makes the <em>ratio close to whole numbers </em>

Multiply by 2. Then

Co:O = 2:2.998 ≈ 2:3

<em>Step 5</em>: Write the <em>empirical formula</em>

EF = Co₂O₃

4 0
3 years ago
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