How to calculate density (what’s the formula) density=_ divided by _

Chemistry

1 answer:

Sphinxa [80]3 years ago

8 0

Answer:

density equals weigt divided by volume

Explanation:

$density = \frac{m}{v}$

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KCl is an ionic salt with a formula weight of 74.54 g/mole. A 1.417 M (mole/liter) KCl solution has a density of 1064.5 grams/li

xenn [34]

Answer:

Mass of the salt: 105.6g of KCl.

Mass water: 958.9g of water.

Molality: 1.478m.

Explanation:

Mass of the salt:

In 1L, there are 1.417 moles. In grams:

1.417 moles KCl * (74.54g / mol) = 105.6g of KCl

Mass of the water:

We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:

1L * (1064.5g / L) = 1064.5g - Mass solution.

Mass water = 1064.5g - 105.6g = 958.9g of water

Molality:

Moles KCl = 1.417 moles KCl.

kg Water = 958.9g = 0.9589kg.

Molality = 1.417mol / 0.9589kg = 1.478m

3 0

3 years ago

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in

LenKa [72]

Answer:

$P=3.7atm$

Explanation:

Hello,

In this case, it is possible to determine the pressures of both helium and neon as shown below:

$n_{He}=\frac{P_{He}V_{He}}{RT}=\frac{5.6atm*3.0L}{0.082\frac{atm*L}{mol*K}*298.15K} =0.688molHe\\\\n_{Ne}=\frac{P_{Ne}V_{Ne}}{RT}=\frac{3.6atm*4.5L}{0.082\frac{atm*L}{mol*K}*298.15K}=0.663molNe$

Now, one considers the total moles (addition between both neon's and helium's moles) and the total volume to compute the final pressure as shown below:

$P=\frac{n_TRT}{V_T} =\frac{(0.688+0.663)mol*0.082\frac{atm*L}{mol*K}*298.15K}{9.0L}=3.7atm$