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2 years ago

The symbol for xenon (Xe) would be a part of the noble gas notation for the element antimony. cesium. radium. uranium.

2 answers:

8 0

The noble gas notation is the short or abbreviated form of the electron configuration.

It means that you use the symbol of the previous noble gas as part of the electron configuration of an element.

The gas noble previous to antimony is Kr, so you do not use Xe to write the electron configuration of Sb.

The gas noble previous to radium is Rn, so you do not use Xe to wirte the electron configuration of Ra.

The gas noble previous to uranium is Rn, so you do not use Xe to write the electron configuration of U.

The gas noble previous to cesium is Xe, so you use Xe to write the noble notation for Sb. This is it: Cs: [Xe] 6s.

Answer: cesium

The ga

atroni [7]2 years ago

7 0

the answer is B.

CESIUM.....

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The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

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

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

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

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

Describe how to prepare the solution using a 250.0 mL volumetric flask by placing the steps in the correct order. Not all of the

padilas [110]

Answer:

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

Explanation:

Preparation of NaCl solution in 250.0 ml volumetric flask:

Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)

After dissolving NaCl add the water upto the mark.

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

6 0

2 years ago

If a gas has a volume of 750 mL at 25oC, what would the volume of the gas be at 55oC?

blagie [28]

Answer:

The answer to your question is V2 = 825.5 ml

Explanation:

Data

Volume 1 = 750 ml

Temperature 1 = 25°C

Volume 2= ?

Temperature 2 = 55°C

Process

Use the Charles' law to solve this problem

V1/T1 = V2/T2

-Solve for V2

V2 = V1T2 / T1

-Convert temperature to °K

T1 = 25 + 273 = 298°K

T2 = 55 + 273 = 328°K

-Substitution

V2 = (750 x 328) / 298

-Simplification

V2 = 246000 / 298

-Result

V2 = 825.5 ml

7 0

2 years ago

A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?

Llana [10]

Let's assume that the gas has ideal gas behavior. 
Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

P = 60 cm Hg = 79993.4 Pa
V = 125 mL = 125 x 10⁻⁶ m³

n = ?

R = 8.314 J mol⁻¹ K⁻¹
T = 25 °C = 298 K

By substitution,
79993.4 Pa x 125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K
n = 4.0359 x 10⁻³ mol

Hence, moles of the gas = 4.0359 x 10⁻³ mol

Moles = mass / molar mass

Mass of the gas = 0.529 g

Molar mass of the gas = mass / number of moles
= 0.529 g / 4.0359 x 10⁻³ mol
 = 131.07 g mol⁻¹

Hence, the molar mass of the given gas is 131.07 g mol⁻¹

4 0

2 years ago

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by pa

MAVERICK [17]

Answer:

Mass percent of nitrogen in the compound is 13,3%

Explanation:

Dumas method is an analytical method to determine nitrogen content in samples, thus:

CₐHₓNₙ + (2a+x/2) CuO → aCO₂ + ˣ/₂ H₂O + ⁿ/₂ N₂ + (2a+x/2) Cu

As the CO₂ is removed with KOH, in the mixture you have H₂O and N₂

At 25°C. the vapor pressure of water is 23,8 torr, that means that the pressure due to N₂ is:

726torr - 23,8torr = 702,2 torr.

Using gas law:

n = PV/RT

Where:

P is pressure (702,2torr≡ 0,924 atm)

V is volume (0,0318L)

R is gas constant (0,082atmL/molK)

And T is temperature (25°C≡298,15K)

Replacing, number of moles of N₂, n, are:

n = 1,20x10⁻³moles of N₂. In grams:

1,20x10⁻³moles of N₂× $\frac{28g}{1mol}$ = 0,0336 g of N₂.

Thus, mass percent of nitrogen in the compound is:

$\frac{0,0336g N}{0,253gSample}$ ×100= 13,3%

I hope it helps!

7 0

2 years ago