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labwork [276]
4 years ago
9

During _______ materials move into the cell while during ______ materials move out of the cell.

Chemistry
1 answer:
Len [333]4 years ago
3 0
D. Endocytosis ; Exocytosis
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Which of the following for an object should be known to calculate its density
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Density = mass / volume

therefore, both mass and volume are needed

the first one represents volume

The Space it occupies is the answer

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Is green ink homogeneous or heterogeneous?
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What do we call living things that share characteristics including processes that make life possible?
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Organisms. Every living thing is an organism and without living organisms, we as a population wouldn’t be able to create a sustainable living environment.:)
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3 years ago
You apply the same amount of heat to five grams of water and five grams of aluminum. The temperature of the aluminum increases m
Mashutka [201]
Specific heat is another physical property of matter. All matter has a temperature associated with it. The temperature of matter is a direct measure of the motion of the molecules: The greater the motion the higher the temperature:



Motion requires energy: The more energy matter has the higher temperature it will also have. Typicall this energy is supplied by heat. Heat loss or gain by matter is equivalent energy loss or gain.

With the observation above understood we con now ask the following question: by how much will the temperature of an object increase or decrease by the gain or loss of heat energy? The answer is given by the specific heat (S) of the object. The specific heat of an object is defined in the following way: Take an object of mass m, put in x amount of heat and carefully note the temperature rise, then S is given by



In this definition mass is usually in either grams or kilograms and temperatture is either in kelvin or degres Celcius. Note that the specific heat is "per unit mass". Thus, the specific heat of a gallon of milk is equal to the specific heat of a quart of milk. A related quantity is called the heat capacity (C). of an object. The relation between S and C is C = (mass of obect) x (specific heat of object). A table of some common specific heats and heat capacities is given below:

Some common specific heats and heat capacities: Substance S (J/g 0C) C (J/0C) for 100 g Air 1.01 101 Aluminum 0.902 90.2 Copper 0.385 38.5 Gold 0.129 12.9 Iron 0.450 45.0 Mercury 0.140 14.0 NaCl 0.864 86.4 Ice 2..03 203 Water 4.179 417.9   

Consider the specific heat of copper , 0.385 J/g 0C. What this means is that it takes 0.385 Joules of heat to raise 1 gram of copper 1 degree celcius. Thus, if we take 1 gram of copper at 25 0C and add 1 Joule of heat to it, we will find that the temperature of the copper will have risen to 26 0C. We can then ask: How much heat wil it take to raise by 1 0C 2g of copper?. Clearly the answer is 0.385 J for each gram or 2x0.385 J = 0.770 J. What about a pound of copper? A simple way of dealing with different masses of matter is to dtermine the heat capacity C as defined above. Note that C depends upon the size of the object as opposed to S that does not.

We are not in position to do some calculations with S and C.

Example 1: How much energy does it take to raise the temperature of 50 g of copper by 10 0C?



Example 2: If we add 30 J of heat to 10 g of aluminum, by how much will its temperature increase?

 



Thus, if the initial temperture of the aluminum was 20 0C then after the heat is added the temperature will be 28.3 0C.
5 0
3 years ago
Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

8 0
3 years ago
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