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3 years ago

Give your familiarity for following terms

Chemistry

1 answer:

KiRa [710]3 years ago

3 0

Answer:

The roasting process is a delicate combination of art and science . Roasters are familiar with how the beans look and the smells Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests .Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests.
Smelting is a process of applying heat to ore in order to extract a base metal. It is a form of extractive metallurgy. It is used to extract many metals from their ores, including silver, iron, copper, and other base metals.In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining.
 In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. ... For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining.
Polling is the process where the computer or controlling device waits for an external device to check for its readiness or state, often with low-level hardware. For example, when a printer is connected via a parallel port, the computer waits until the printer has received the next character.

Explanation:

hope it heloed

You might be interested in

To grow or reach the next stage in a life cycle is to develop.

gogolik [260]

Answer:

It's obviously true

Explanation:

As we have evolved over the years we have become more advanced

5 0

3 years ago

Give an example of a time when you would want to increase friction.

Snezhnost [94]

1) you want to increase friction when it gets cold. If you're outside and it's really cold, you're going to rub your hands to warm them up, therefore friction is increasing

I'm not do sure about decreasing.

6 0

3 years ago

a piece of food is burned in a calorimeter that contains 200.0g of water. If the temperature of the water rose from 65.0°C to 83

Flauer [41]

Answer: 15062.4 Joules

Explanation:

The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of food = 200.0g

C = 4.184 j/g°C

Φ = (Final temperature - Initial temperature)

= 83.0°C - 65.0°C = 18°C

Then, Q = MCΦ

Q = 200.0g x 4.184 j/g°C x 18°C

Q = 15062.4 J

Thus, 15062.4 joules of heat energy was contained in the food.

4 0

4 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

3 years ago

What do radio waves and gamma rays have in common?

padilas [110]

Answer: The sequence from longest wavelength (radio waves) to shortest wavelength (gamma rays) is also a sequence in energy from lowest energy to highest energy. ... The energy carried by a radio wave is low, while the energy carried by a gamma ray is high. Different materials can block different types of light.

PLEASE MARK BRAINLIEST

5 0

3 years ago

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