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Jet001 [13]
3 years ago
14

Analysis of an unknown substance indicates that it is a brittle solid. It acts as an electrical insulator (does not conduct) as

a solid but conducts electricity when melted. Which of the following substances would have these characteristics?
a. HCl
b. Al
c. KBr
d. SiF4
e. I2
Chemistry
1 answer:
Musya8 [376]3 years ago
5 0

Answer:

KBr

Explanation:

A metal has a very high melting point. Aluminum is a metal.

Ionic solids do not conduct electricity in the solid state. However, when the solids are melted, they conduct electricity.

KBr is an ionic substance, it can only conduct electricity in molten form because the ions are locked up in the solid.

You might be interested in
22.17 grams of Mercury (II) Nitrate, Hg(NO3)2, reacts with an excess of Potassium, K.
Svetradugi [14.3K]

Answer:

Mass = 14.0 g

Explanation:

Given data:

Mass of mercury nitrate = 22.17 g

Mass of mercury formed = ?

Solution:

Chemical equation:

Hg(NO₃)₂ + 2K    →    2KNO₃ + Hg

Number of moles of mercury nitrate:

Number of moles = mass/molar mass

Number of moles = 22.17 g / 324.6 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of Hg(NO₃)₂ and mercury.

                Hg(NO₃)₂       :        Hg

                     1                :          1

                  0.07             :       0.07

Mass of mercury:

Mass = number of moles × molar mass

Mass = 0.07 mol × 200.6 g/mol

Mass = 14.0 g

7 0
2 years ago
5. A 5.00 g sample of an unknown substance was heated from 25.2 C to 55.1 degrees * C , and it required 133 to do so. Identify t
Leviafan [203]

Answer:

Aluminum

Explanation:

Given

T_1 = 25.2^oC

T_2 = 55.1^oC

m = 5.00g

\triangle Q= 133J

<em>See attachment for chart</em>

Required

Identify the unknown substance

To do this, we simply calculate the specific heat capacity from the given parameters using:

c = \frac{\triangle Q}{m\triangle T}

This gives:

c = \frac{\triangle Q}{m(T_2 - T_1)}

So, we have:

c = \frac{133J}{5.00g * (55.1C - 25.2C)}

c = \frac{133J}{5.00g * 29.9C}

c = \frac{133J}{149.5gC}

c = 0.89\ J/gC

From the attached chart, we have:

Al(s) = 0.89\ J/gC --- The specific heat capacity of Aluminum

<em>Hence, the unknown substance is Aluminum</em>

7 0
2 years ago
A plastic 2L soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an emp
Aneli [31]

Answer:  The new volume be if you put it in your freezer is 1.8 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=2L\\T_1=25^oC=(25+273)K=298K\\V_2=?\\T_2=-4.0^oC=((-4)+273)K=269K

Putting values in above equation, we get:

\frac{2}{298}=\frac{V_2}{269}\\\\V_2=1.8L

The new volume be if you put it in your freezer is 1.8 L

8 0
2 years ago
Convert 7.50 grams of glucose C6H12O6 to moles
Lera25 [3.4K]

Answer:

The number of mole is 0.04167mole

Explanation:

To convert gram to mole, we need to calculate the molecular weight of the compound

C6H12O6

C - 12

H - 1

O - 16

Molecular weight = 6 * 12 + 1 *12 + 6 * 16

= 72 + 12 + 96

= 180g/mol

To covert gram to mole

Therefore,

= 7.50g/ 180g/mol

= 0.04167 mole of glucose

3 0
3 years ago
What is the mass of 0.714 moles of Mercury (I) Chloride (Hg2Cl2)?
ArbitrLikvidat [17]
Data:

m (<span>Sample Mass) = ? 
n (</span><span>Number of moles) = 0.714 mol
MM (Molar Mass) of </span>Mercury (I) Chloride (Hg_{2}  Cl_{2})
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass Hg_{2} Cl_{2} = 401.18 + 70.906 = 472.086 ≈ 472.09<span> amu or 472.09 g/mol
</span>
Formula:

n =  \frac{m}{MM}

Solving:


n = \frac{m}{MM}
0.714 =  \frac{m}{472.09}
m = 337.07226\:\to\:\boxed{\boxed{m\approx 337 g}} \end{array}}\qquad\quad\checkmark

Answer:
By approximation would be letter D) <span>337.2 g</span>


8 0
3 years ago
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