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3 years ago

Decide whether the compound is ionic or molecular, if you can.

Chemistry

1 answer:

mestny [16]3 years ago

7 0

Answer:

Molecular solid

Explanation:

A molecular solid has a low melting point, they are soft and do not conduct electricity.

We have been told in the question that the solid does not really dissolve in water and it's solution does not improve the electrical conductivity of water. Hence, it must be a molecular solid.

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The symbol for xenon (xe) would be a part of the noble gas notation for the element

daser333 [38]

Answer:

Explanation:

To write the electronic configuration of an element using the preceding noble gas configuration, we simply use the noble gas in the previous period of that particular element. We do not use the noble gas of that period in which the element belongs to but the one in the preceding group.

A is wrong

This is because the noble gas in the preceding period is krypton.

In fact, the electronic configuration is [Kr] 4d105s25p3

C is wrong

The noble gas in the last period here is Radon. The electronic configuration of Radium in fact is [Rn] 7s2

D is wrong

The noble gas in the last period before the period of uranium is Radon also. In fact, the electronic configuration of the element is [Rn] 5f36d17s2

B is correct, the actual electronic configuration of the element Cesium is [Xe] 6s1

6 0

3 years ago

Oxidation unit test

3241004551 [841]

In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.

Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.

Next, we can write the following $x$ , since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

$Mn^xI_2^-\\\\Mn ^xO_2^{-2}$

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

$Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4$

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.

Learn more:

8 0

3 years ago

a gas thermometer measure temperature by measuring the pressure of a gas inside the fixed volume container a thermometer reads a

sdas [7]

Hello!

a gas thermometer measure temperature by measuring the pressure of a gas inside the fixed volume container a thermometer reads a pressure of 248 torr at 0 degrees Celsius what is the temperature when the thermometer reads a pressure of 345 torr

We have the following information:

P1 (initial pressure) = 248 torr

T1 (initial temperature) = 0 ºC (in Kelvin)

TK = TºC + 273.15 → TK = 0 + 273.15 → T1 (initial temperature) = 273.15 K

P2 (final pressure) = 345 torr

T2 (final temperature) = ? (in Kelvin)

According to the Law of Charles and Gay-Lussac in the study of gases, we have an isochoric (or isovolumetric) transformation when its volume remains constant or equal, then we will have the following formula:

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

$\dfrac{248}{273.15} = \dfrac{345}{T_2}$

$248*T_2 = 345*273.15$

$248\:T_2 = 94236.75$

$T_2 = \dfrac{94236.75}{248}$

$T_2 = 379.9868952... \to \boxed{\boxed{T_2 \approx 380\:K}}\end{array}}\qquad\checkmark$

If you want the solution in Celsius Temperature, we have:

TC = TK - 273.15

TC = 380 - 273.15

TC = 106.85 → Temperature = 106.85 ºC

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

6 0

3 years ago

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate

vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

k = Ae^-(Ea/RT)

where, k = rate constant

A = pre-exponential factor

Ea = activation energy

R = gas constant

T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

T1= (244 + 273.15) K = 517.15 K

T2= (324+ 273.15) K =597.15 K

K1 = 6.7 , K2 = ?

R = 8.314 J/mol K

Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

7 0

3 years ago

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A judge in a juvenile court can dispose a juvenile runaway to a residential facility.

ratelena [41]

The answer is true. I know this because it almost happened to me.

3 0

3 years ago

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