According to Graham's law of effusion, the ratio of the rates of effusion of two different molecules is equal to the square root of the reciprocal of the molar masses of the molecules. gas a is faster than B since it is lighter than B. Hence we expect a molar mass less than 17. we have the equation 1/0.68 =
where x is the molar mass of gas A. This is equal to 7.86 grams.
Answer:
0.4762 J/g°C.
Explanation:
<em>The amount of heat released to water = Q = m.c.ΔT.</em>
where, m is the mass of water (m = 15.0 g).
c is the specific heat capacity of water = ??? J/g°C.
ΔT is the temperature difference = (final T - initial T = 37.0°C - 30.0°C = 7.0°C).
<em>∴ The specific heat capacity of water = c = Q/m.ΔT</em> = (50.0 J)/(15.0 g)(7.0°C) = <em>0.4762 J/g°C.</em>
They also have a specialized non-lignified tissue (the phloem) to conduct products of photosynthesis. Vascular plants include the clubmosses<span>, </span>horsetails<span>, </span>ferns,gymnosperms<span> (including </span>conifers<span>) and </span>angiosperms<span> (flowering plants). Scientific names for the group include </span>Tracheophyta<span> and </span>Tracheobionta<span>.</span>
Part 1 : Answer is only B substance is soluble in water.
In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.
Part 2 : Substance A is not soluble in water and substance B is soluble in water.
According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.
But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.