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QveST [7]
3 years ago
6

describe how a physical property such as mass or Texture can change without causing a change in the substance

Chemistry
1 answer:
Tom [10]3 years ago
3 0

You have to be very careful with this question. A change in mass can also occur in chemical changes especially if you have too much of something. For example

CH4 + 1.5 02 ===> CO2 + H2O

If you have too much of either CH4 or O2, there will be some CH4 or O2 left over. There has been a change in mass that you have too much of.

However that is not the point of the question. It is just something you need to be aware of.

Suppose you have a piece of aluminum and you take a course grinder after it. You will change the texture of the side you took the grinder to. If the aluminum has been anodized (a color has been put on it's surface), you may grind the color off or if it is just plain aluminum, you may roughen the surface, but you won't change what the aluminum will do chemically.

You may need only a small portion of the aluminum and you grind off just what you need. That will change the mass of both what you took off and the piece that you want, but the aluminum will still do whatever chemical property you need to use.

So you can change both texture and mass without changing the chemical properties of the substance whose mass or texture you are changing.

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Convert 1.25 kg into grams
umka2103 [35]
<h2>1 kilogram = 1,000 grams</h2><h2></h2>

Since we are going from a larger unit to a smaller unit, we multiply.

So we multiply 1.25 by the conversion factor, which is 1,000.

This gives us 1,250 grams

7 0
3 years ago
A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pre
Goryan [66]

Answer:

\boxed {\boxed {\sf 82.7 \textdegree C}}

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

\frac {P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58  \ atm *2.46 \ m^3}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then multiply the first denominator by the second numerator.

(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

The units of atmospheres and cubic meters cancel.

T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}

Solve inside the parentheses.

T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}

T_2= \frac{369.246}{4.4625} \textdegree C}

T_2 = 82.74420168 \textdegree C

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

T_2 \approx 82.7 \textdegree C

The temperature is approximately <u>82.7 degrees Celsius.</u>

3 0
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Answer:

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Explanation:

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How many atoms are found in 12.0 grams of Helium?
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<span>Each mole contains Avagodro's number of atoms i.e. 6.023x10^23, so

3 moles x 6.023x10^23 atoms/mole = 18.069x10^23 atoms = 1.8x19^24 atoms </span>
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