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4 years ago

Which characteristic(s) makes ionic compounds useful as a building material?

2 answers:

8 0

The answer is hardness and brittleness. The ion discharging procedure and taking after impact on reservoir water quality file, for example, hardness, pH, conductivity, has been dissected. Comes about uncovered that the most discharge quality of various materials was touching base at the 30s after startup.

NISA [10]4 years ago

7 0

The correct answer to the question is hardness and brittleness.

EXPLANATION;

Before coming into any conclusion, first we have to understand the formation of an ionic compound.

An ionic compound is generally formed between electropositive metal and electronegative nonmetal.

The electropositive metal will lose electrons and forms a cation . The electronegative non metal will accept electrons and forms anion. The cation and anion so formed will attract each other due to the coulombic force of attraction which results into the formation of an ionic compound.

Due to the strong electrostatic force of attraction between ions, the melting and boiling points of ionic compounds are very high. The ionic compounds have very hardness, and are brittle in nature due to this property of the ionic compound.

As water is a polar compound, so ionic compounds are dissolved in water. Ionic compounds are neutral in nature, but they conduct electricity when they are dissolved in water.

As ionic compounds have hardness, and are brittle in nature, it can be considered as the building material.

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Deffense [45]

Explanation:

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3 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

3 0

4 years ago

An avalanche takes place when ice and snow move down a slope with great speed. What effect does friction have on its rate of mov

vredina [299]

The answer is; D

The friction causes the ice at the base of the avalanche to melt into water (also due to the pressure of the weight of the ice-rock above). The melted water acts as a lubricant hence reducing the drag/friction in the avalanche movement. This increases its speed down the slope hence making it more destructive.

5 0

4 years ago

Read 2 more answers

Two identical steel balls, each of mass 67.8 g, are moving in opposite directions at 4.80 m/s.They collide head-on and bounce ap

xz_007 [3.2K]

Answer:

Explanation:

To detrmine the time interval at which the balls are in contact.

<u>Given information</u>

The mass of each steal ball 67.8g. The speed of ball towards each other is 4.80 m/s. The exerted force by each jaw 15.9 kN and the forece reduce the diameter by 0.130 mm.

Expression for the effective spring constant ball is shown below.

K = |F|/|x|

Here,

k is a spring constant

F is the force exerted on the ball

x is dispalcement due force

substitute 15.9 kN for F and 0.130 mm in above equation

The spring constant is 122 x 10⁶ N/m

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3 years ago

Mr .hitch taught us about sedimentary, metamorphic,and igneous rocks. He describes how they were formed , what they contain, and

Feliz [49]

The answer is C) Geologist

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3 years ago