Answer:2.0M
Explanation
The density of gold is 19.32. The two reasons due to which the statement is incomplete are there is no unit and it doesn't present the temperature and pressure.
Density has been defined as the mass per unit volume it means that mass is present in one meter cube.
S.I unit of density is kg/meter^3.
Mathematically
Density = Mass/Volume.
Momentum has been the product of mass and velocity. It is a vector quantity and it's S.I unit kg x meter/sec. There will be no effect of the outside environment on the potential energy that has to be present inside the body. The kinetic energy decrease or increase with change in the velocity.
Weight has been defined as the total heaviness of the body or object. It has been also known as the portion of anything weighs. In physics weight has been known as the vertical force which has experienced by mass which is due to the gravitation. The weight has been equal to the mass of the body and the product of mass and acceleration of free fall is known as weight.
Therefore, The density of gold is 19.32. The two reasons due to which the statement is incomplete are there is no unit and it doesn't present the temperature and pressure.
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Using Charles's law, the calculation indicates a decrease in temperature by about 146.58 degrees Celsius when a 2.00 L volume of gas at 20.0 degrees celsius is compressed to 1.00 L.
To calculate the decrease in temperature due to compression, we can utilize Charles's Law, provided we are ignoring any changes in pressure. Initially, we have a volume (V₁) of 2.00 L at a temperature (T₁) of 20.0 degrees celsius which is equivalent to 293.15 K (converting to Kelvin). The volume is then compressed to 1.00 L (V₂) and we need to find the new temperature (T₂).
According to Charles' Law (V₁/T₁ = V₂/T₂), after we input the values from the problem, we can isolate T₂, resulting in following equation: T₂ = (V₂* T₁)/ V₁. Substituting the given values, we get T₂ = (1.00L*293.15 K)/2.00 L = 146.57 K.
However, we want to find the decrease in temperature in celsius. Difference in temperature in Celsius and Kelvin scales are the same. So, the decrease in temperature (ΔT) = T₁ - T₂ = 20.0°C - (-126.58°C) = 146.58°C. So, when the volume of the gas is compressed from 2.00 L to 1.00 L, the temperature decreases by about 146.58 degrees Celsius.
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Using the combined gas law, we find that the new pressure of the gas sample, after it's compressed and heated, is approximately 353.8 kPa.
The subject of this question is gas laws, specifically the combined gas law which states that the ratio of the product of pressure and volume and the absolute temperature of a gas is constant. We apply this law to calculate the new pressure of the gas sample. Starting from the conditions of STP (Standard Temperature and Pressure, defined as 273.15 K and 1 atm i.e., 101.325 kPa), the volume of gas is decreased from 700.0 mL to 200.0 mL and the temperature is increased from 273.15 K to 30.0 degrees Celsius (or 303.15 K in absolute terms).
We set up the equation P1*V1/T1 = P2*V2/T2, where P1 = 101.325 kPa, V1 = 700.0 mL, T1 = 273.15 K, V2 = 200.0 mL, and T2 = 303.15 K. Plugging in these numbers and solving for P2 (the new pressure), we get P2 = P1*V1*T2 / (T1*V2) = 101.325 kPa * 700.0 mL * 303.15 K / (273.15 K * 200.0 mL) = approximately 353.8 kPa.
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