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3 years ago

How is the miscibility of two liquids related to their polarity?

Chemistry

1 answer:

makvit [3.9K]3 years ago

4 0

Answer:

The more polar the liquid, the more likely that it is miscible with water. The polarity of a liquid does not affect its miscibility with water. The less polar the liquid, the more likely that it is miscible with water. The more polar the liquid, the less likely that it is miscible with water.

Explanation:

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An irregularly-shaped piece of copper (Cu) has a mass of 55.0 grams. What is the volume in cm³ of this piece of copper if its de

vredina [299]

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question we have

$volume = \frac{55}{8.96} \\ = 6.138392...$

We have the final answer as

Hope this helps you

8 0

3 years ago

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

posledela

The question is incomplete, here is the complete question:

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

$2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag$

Suppose 6 moles of silver nitrate react. The reaction consumes___ moles of copper. The reaction produces __ moles of copper(II) nitrate and __ moles of silver.

<u>Answer:</u> The amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

<u>Explanation:</u>

We are given:

Moles of silver nitrate = 6 moles

For the given chemical reaction:

$2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag$

<u>For copper metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of copper metal

So, 6 moles of silver nitrate will react with = $\frac{1}{2}\times 6=3mol$ of copper metal

Moles of copper reacted = 3 moles

<u>For copper(II) nitrate:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 1 mole of copper(II) nitrate

So, 6 moles of silver nitrate will produce = $\frac{1}{2}\times 6=3mol$ of copper(II) nitrate

Moles of copper(II) nitrate produced = 3 moles

<u>For silver metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6 moles of silver nitrate will produce = $\frac{2}{2}\times 6=6mol$ of silver metal

Moles of silver metal produced = 3 moles

Hence, the amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

6 0

4 years ago

What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S

Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

<u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

c= Heat Capacity of Liquid= 4.184 $\frac{J}{gC}$
m= 185.5 g
ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 $\frac{J}{gC}$ × 185.5 g× (- 25.6 °C)

Solving:

<u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× $\frac{1mol}{18 grams}$ = 10.30 moles, where 18 $\frac{g}{mol}$ is the molar mass of water, that is, the amount of mass that a substance contains in one mole.