Answer:
Nylon and Spandex (Lycra).
Explanation:
These materials are designed to fit the body, with nylon drying more quickly than other types of fabrics, and Spandex being commonly found in swimming and sports apparel due to its elastic qualities. Both fabrics also wick moisture away and dry rapidly.
With high capacity and enhanced flexibility, nylon and Spandex provide a snug fit to the body and can retain their shape during various activities, making them ideal for swimming.
This explains why these materials are suitable based on the situation given.
Answer:
9.69g
Explanation:
To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.
1 mole of gas takes up 22.4L at STP.
Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning
X moles of N2 = 7.744/22.4 = 0.346 moles
Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:
Molar Mass of N2 = 2x14 = 28g/mol
Number of moles N2 = 0.346 moles
Find the mass of N2 =?
Mass = number of moles × molar mass
Mass of N2 = 0.346 × 28
Mass of N2 = 9.69g
Hence, 7.744L of N2 consists of 9.69g of N2
<span>Some solutions demonstrate colligative properties, which rely on the quantity of solute in a solvent. To find the elevation in boiling point, we use the formula:
</span><span>ΔT(boiling point) =
(Kb)mi
where Kb represents a constant, m is the solution's molality, and i is the van't Hoff factor.
From the provided information, we can easily determine i as follows:
</span>ΔT(boiling point) = (Kb)mi
103.45 - 100 = (0.512)3.90i
i = 1.73 <-------van't Hoff factor
Specific heat refers to the quantity of heat a material can absorb or release to alter its temperature by one degree Celsius. To calculate specific heat, we apply the equation for the heat absorbed by the system. The heat taken in or released by a system can be expressed by multiplying the mass of the substance by its specific heat capacity and the change in temperature. The formula is:
Heat = mC(T2-T1)
By substituting the provided values, we can find C, the specific heat of the substance.
2510 J = 0.158 kg (1000 g / 1 kg)(C)(61.0 - 32.0 °C) C = 0.5478 J/g°C
<span>128 g/mol
Applying Graham's law of effusion, we can utilize the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = effusion rate of gas 1
r2 = effusion rate of gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998.
We can now insert the known values into Graham's equation to find m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
Thus, we find m2 to be 127.992.
Rounding to three significant figures yields 128 g/mol</span>
