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lozanna [386]
4 years ago
10

What is the density of an object that has a mass of 119 grams and a volume of 30 ml?

Chemistry
1 answer:
zimovet [89]4 years ago
8 0
Density equals mass divided by volume.
119/30 equals 3.97 g/cc (rounded)
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A sample of oxygen gas at a pressure of 1.19 atm and a temperature of 24.4 °C, occupies a volume of 18.7 liters. If the gas is a
mihalych1998 [28]

Answer:

\boxed {\boxed {\sf 0.757 \ atm}}

Explanation:

We are asked to find the pressure of a gas given a change in volume. Since the temperature remains constant, we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume is inversely proportional to the pressure. The formula for this law is:

P_1V_1= P_2V_2

Initially, the oxygen gas occupies a volume of 18.7 liters at a pressure of 1.19 atmospheres.

1.19 \ atm * 18.7 \ L = P_2V_2

The gas expands to a volume of 29.4 liters, but the pressure is unknown.

1.19 \ atm * 18.7 \ L = P_2 * 29.4 \ L

We are solving for the new pressure, so we must isolate the variable P_2. It is being multiplied by 29.4 liters. The inverse operation of multiplication is division. Divide both sides of the equation by 29.4 L.

\frac {1.19 \ atm * 18.7 \ L}{29.4 \ L} =\frac{ P_2 * 29.4 \ L}{29.4 \ L}

\frac {1.19 \ atm * 18.7 \ L}{29.4 \ L} =P_2

The units of liters cancel.

\frac {1.19 \ atm * 18.7 }{29.4 } =P_2

\frac {22.253}{29.4 } \ atm = P_2

0.7569047619 \ atm =P_2

The original measurements all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 9 in the ten-thousandth place to the right of this place tells us to round the 6 up to a 7.

0.757 \ atm \approx P_2

The pressure of the gas sample is approximately <u>0.757 atmospheres.</u>

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3 years ago
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Which is a property that can be used to identify matter?
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Answer is All of the above
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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.
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Superstitions :

1. Gazing too long at the full moon will bring out one’s inner lunatic.

2. The waxing of a new moon is the best time for sowing seeds, weddings, childbirth, new projects and travel.

3. It is bad luck to see the new moon, for the first time of the night, through glass or tree branches.

Hope this helps and if it does, don't be afraid to give my answer a "Thanks" and maybe a Brainliest if it's correct?  

6 0
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