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3 years ago

Which element consists of positive ions immersed in a “sea” of mobile electrons?

Chemistry

1 answer:

denis-greek [22]3 years ago

8 0

Answer:

Calcium

Explanation:

Calcium is the only metal here

hope this helps...

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erik [133]

The answer is A, you winn not find carbon dioxide in ur pee lol I hope this helps out, have an amazing day!!

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3 years ago

Read 2 more answers

How can two objects have the same temperature but different thermal energies?

Mamont248 [21]

Thermal energy is dependent on the mass. If objects have a different mass then they will have a different thermal energies.

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3 years ago

PLSSSSS HELP I DONT GET THIS PROBLEMMMM

Aleks [24]

Answer:

C. 7370 joules.

Explanation:

There is a mistake in the statement. Correct form is described below:

<em>Using the above data table and graph, calculate the total energy in Joules required to raise the temperature of 15 grams of ice at -5.00 °C to water at 35 °C. </em>

The total energy needed to raise the temperature is the combination of latent and sensible heats, all measured in joules, and represented by the following model:

$Q = m\cdot [c_{i} \cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})]$ (1)

Where:

$m$ - Mass of the sample, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of the sample, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of water, in degrees Celsius.

$Q$ - Total energy, in joules.

If we know that $m = 15\,g$ , $c_{i} = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334.72\,\frac{J}{g}$ , $T_{1} = -5\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ and $T_{3} = 35\,^{\circ}C$ , then the final energy to raise the temperature of the sample is:

$Q = (15\,g)\cdot \left[\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (5\,^{\circ}C)+ 334.72\,\frac{J}{g} + \left(4.184\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (35\,^{\circ}C) \right]$

$Q = 7371.9\,J$

Hence, the correct answer is C.

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3 years ago

Alexa pushed a cart against the wall with 500 newtons of force. The cart didn't move.

liraira [26]

Hopefully this helps sorry

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3 years ago

Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure

Galina-37 [17]

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

where,

$\DeltaT_b$ = change in freezing point

$k_b$ = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of $KCl$ will be,

$KCl\rightarrow K^++Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M $KCl(aq)$ at standard pressure, the freezing point of 1.0 M $CaCl_2(aq)$ at standard pressure is lower.

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3 years ago