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3 years ago

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Two liquids, a and b are immiscible. liquid a has a density of 0.89 g/ml. liquid b has a density of 0.72 g/ml. what would you ex

pect to see if you added equal amounts of both liquids to a test tube, shook, and let settle

1 answer:

Burka [1]3 years ago

5 0

In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.

The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.

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A 6.00 L container of N_2 has a temperature of 273 K. Calculate the volume if the temperature is doubled.

vfiekz [6]

V1/T1 =V2/T2 (using charles law)

V1=6.00
V2=?
T1=273
T2=273

Making V2 the subject of formula the equation then becomes

V2= V1xT2/T1

6.00x263/273=6.0L

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2 years ago

The North Pole of a magnet will attract to the South Pole of another magnet. True or False

hodyreva [135]

pretty sure its True hope this helped

Explanation:

A magnet has two ends called poles; one end is the north pole and the other is the south pole. A north pole will attract a south pole; the magnets pull on each other. But the two north poles will push each other away. ... The magnet is attracted by the earth's magnetic north pole and always points in that direction.

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2 years ago

Select the correct answer from each drop-down menu. consider the substances hydrogen (h2), fluorine (f2), and hydrogen fluoride

nasty-shy [4]

The boiling point of HF is higher than the boiling point of $H_2$ , and it is higher than the boiling point of $F_2$ .

<h3>What is the boiling point?</h3>

The boiling point is the temperature at which the pressure exerted by the surroundings upon a liquid is equalled by the pressure exerted by the vapour of the liquid.

$F_2$ has weak dispersion force attractions between its molecules, whereas liquid HF has strong ionic interactions between $H^+$ and $F^-$ ions.

Only London Forces are formed - Therefore more energy is required to break the intermolecular forces in HF than in the other hydrogen halides and so HF has a higher boiling point.

$H_2$ and $F_2$ will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.

Hence, the boiling point of HF is higher than the boiling point of $H_2$ , and it is higher than the boiling point of $F_2$ .

Learn more about the boiling point:

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1 year ago

How many gases are there that are not noble gases

natulia [17]

Answer:

Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

Explanation:

I pretty much covered it in my answer!

Pls Brainliest! It would mean a lot! ;)

8 0

3 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago