An element had five valence electrons for bonding. This element is most likely which of the following?

A natural sample of boron consists of two isotopes

amid [387]

Boron: isotope data. Both isotopes ofBoron, B-10 and B-11, are used extensively in the nuclear industry. B-10 is used in the form of boric acid as a chemical shim in pressurized water reactors while in the form of sodium pentaborate it is used for standby liquid control systems in boiling water reactors

7 0

3 years ago

If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml

Bumek [7]

Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

$\frac{V}{T} =k$

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

$\frac{P}{T} =k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T} =k$

Having two different states, an initial state and an final state, it is true:

$\frac{P1*V1}{T1} =\frac{P2*V2}{T2}$

In this case:

P1= 0.9 atm
V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

P2= 1 atm
V2= ?
T2= 0°C= 273 °K

Replacing:

$\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}$

Solving:

$V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}$

V2= 2.415 L =2,415 mL

The new volume is 2415 mL

6 0

3 years ago

Atmospheric chemistry involves highly reactive, odd-electron molecules such as the hydroperoxyl radical HO2, which decomposes in

dem82 [27]

Answer:

Rate = k [HO2]

Rate constant = 0.8456us-1

Explanation:

Time(us) 0 0.6 1.0 1.4 1.8 2.4

[HO2](uM) 8.5 5.1 3.6 2.6 1.1 1.1

The rate law is given as;

Rate = k [HO2]^x

Where x signify the order of reaction.

For an order of reaction, the rate constant is constant for all concentrations. We are going to use this to obtain the order of reaction.

Zero Order:

[A] = [A]o -kt

5.1 = 8.5 - k(0.6)

-k (0.6) = 5.1 - 8.5

k = 5.67

3.6 = 5.1 - k(0.4)

-k (0.4) = 3.6 - 5.1

k = 3.75

The fact that the rate constant was not constant means the reaction is not a zero order reaction.

First Order:

ln[A] = ln[A]o -kt

(5.1) = ln(8.5) - k(0.6)

-k (0.6) = ln(5.1) - ln(8.5)

k = 0.8524

ln(3.6) = ln(5.1) - k(0.4)

-k (0.4) = ln(3.6) - ln(5.1)

k = 0.8708

ln(2.6) = ln(3.6) - k (0.4)

-k (0.4) = ln(2.6) - ln(3.6)

k = 0.8136

From the three calculations we see that the value of the rate constant is fairly constant in the range of 0.8 This means our reaction is a first order reaction.

The rate law is given as;

Rate = k [HO2]

We can represent the rate constant as the average of the three rate constants calculated above;

Rate constant = (0.8136 + 0.8708 + 0.8524 / 3)

Rate constant = 0.8456us-1

6 0

3 years ago

What is the distinction between the caus and the manner of death

kompoz [17]

The manner of death is how the death came about. A natural death occurs as a result of aging, illness, or disease. ... A homicide occurs when death is caused by another person.

5 0

3 years ago

Cuál o cuáles de las siguientes características son correctas para la mayoría de los compuestos del carbono?

garik1379 [7]

Answer:

Todas son correctas

Explanation:

I) Los compuestos del carbono son solubles en disolventes no polares como el hexano, benceno, éter, etc. VERDADERO. La mayoría de los compuestos de carbono son apolares, y basados en la regla: Similar disuelve similar, podemos presumir que la mayoría de compuestos de carbono se disuelven en solventes no polares.

II) Los compuestos del carbono generalmente presentan puntos de fusión y de ebullición bajos. VERDADERO. Al ser sustancias apolares, sus fuerzas electrostáticas son bajas. De la misma manera, como su masa atómica es pequeña, las fuerzas de Van der Waals son despreciables haciendo que sus puntos de fusión y ebullición sean bajos respecto a sustancias de estrucutra similar.

III) La gran mayoría de los compuestos del carbono son combustibles, sean estos, gaseosos, líquidos o sólidos. VERDADERO. Los gases (Como el gas natural) son combustibles usados para cocina. Los líquidos (Como la gasolina) son combustibles y bastante inflamables. Los sólidos (Como la madera) se usan como combustibles para hacer asados o son el combustible en incendios forestales.

IV) El enlace covalente es característico de los compuestos del carbono. VERDADERO. La polaridad del carbono es neutral haciendo que la mayoría de los enlaces que forma sean covalentes.

<h3>Todas son correctas </h3>

5 0

3 years ago