An increase in the concentration of sugar in a solution leads to the increase in boiling point of the solution.
<h3>What is boiling point?</h3>
Boiling point of a solution is that pressure exerted on a liquid by its surrounding is equal to the vapour pressure of the liquid.
When there is an increase in concentration of solute, more space will be occupied in the fluid leading to an increase in the boiling point of the fluid.
Therefore, An increase in the concentration of sugar in a solution leads to the increase in boiling point of the solution.
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Answer:
d = 1.24 kg/m³
v = 0.81 m³/kg
Explanation:
To do this, we need to analyze the given data and know the expressions we need to use here to do calculations.
We have a pressure of 1.05 atm and 300 K of temperature. To determine the density, we need to use a similar expression of an ideal gas. In this case, instead of using moles, we will use density:
P = dRT
d = P/RT (1)
Where:
R: universal constant of gases
d: density.
From here we can determine the specific volume by using the following expression:
v = 1/d (2)
Now, as we are looking for density, we need to convert the units of pressure in atm to Pascal (or N/m) and the conversion is the following:
P = 1.05 atm * 1.013x10⁵ N/m atm = 106,365 N/m
Now, using R as 287 the density would be:
d = 106,365 / (287 * 300)
<h2>
d = 1.24 kg/m³</h2>
Finally the specific volume:
v = 1 / 1.41
<h2>
v = 0.81 m³/kg</h2>
Hope this helps
Heat can be reduced but it cannot be eliminated completely there is some amount if heat energy present even if it is absolute zero(-273degrees Celsius)
Answer:
2500 kg/m³
Explanation:
The density is ...
(107 g)/((12.2 cm)(3.7 cm)(0.95 cm)) = 107/42.883 g/cm³ ≈ 2.5 g/cm³
In kg/m³, the density is ...
(2.5 g/cm³)×(1 kg/(1000 g))×(100 cm/m)³ = 2500 kg/m³
The conversion from CGS to MKS units of density is done using the usual method of units conversion. The conversion factor multipliers each change the units without changing the value, because their numerator is equal to the denominator.
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2 significant figures are used because the block dimensions have a precision of 2 significant figures.
Samburu's weight is (mass x gravity) = (1650kg x 9.8 m/s²) = 16,170 Newtons.
Samburu's contact area with the ground is (0.25 m²/hoof) x (4 hoofs) = 1 m² .
Pressure = (force) / (area) = (16,170 Newtons) / (1 m²) = 16,170 Pascal .
