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ruslelena [56]
3 years ago
12

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

Chemistry
1 answer:
daser333 [38]3 years ago
6 0

Answer:

B

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

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What is the oxidation state of S in HSO4^2-? (A.P.E.X)
BlackZzzverrR [31]

I think there is a typo because I've never seen HSO4 2- before in my life. It should be HSO4-. For that, H is 1+ and each Oxygen is 2-0 totaling 8-. So the oxidation state of sulfur +1 - 8 = 7

So the oxidation state of sulfur is +6

7 0
3 years ago
A bathtub is filled with 100 L of water. A cat jumps in and the volume rises to 125 L. If the cat has a mass of
kirza4 [7]

Answer:

<h2>The answer is 3.2 g/L</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of cat = 400 g

volume = final volume of water - initial volume of water

volume = 125 - 100 = 25 L

The density of the cat is

density =  \frac{400}{125}  =  \frac{16}{5}  \\

We have the final answer as

<h3>3.2 g/L</h3>

Hope this helps you

3 0
3 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (
Mnenie [13.5K]

Answer:

3.1 kg

Explanation:

Step 1: Write the balanced combustion equation

C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O

Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.

The molar mass of C₈H₁₈ is 114.23 g/mol.

1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol

Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈

The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.

Step 4: Calculate the mass corresponding to 70 moles of CO₂

The molar mass of CO₂ is 44.01 g/mol.

70 mol × 44.01 g/mol = 3.1 × 10³ g = 3.1 kg

5 0
3 years ago
A gas mixture contains 10.0 mole% H2O (v) and 90.0 mole % N2. The gas temperature and absolute pressure at the start of each of
Rainbow [258]

Answer: (a). T = 38.2 °C     (b). V = 1.3392 cm³     (c). ii and iii  

Explanation:

this is quite easy to solve, i will give a step by step analysis to solving this problem.

(a). from the question we have that;

the Mole fraction of Nitrogen, yи₂ = 0.1

Also the Mole fraction of Water, yн₂o = 0.1

We know that the vapor pressure is equal to the partial pressure because the vapor tends to condense at due point.

ρн₂o = ṗн₂o

      = yн₂oP = 0.1 × 500 mmHg = 50 mmHg

from using Antoine equation, we apply the equation

logρн₂o = A - B/C+T

T = B/(A - logρн₂o) - C

  = 1730.63 / (8.07131 - log 50mmHg) - 223.426

T = 38.2 °C

We have that the temperature for the first drop of liquid form is 38.2 °C

(b). We have to calculate the total moles of gas mixture in a 30 litre flask;

   n  = PV/RT  

   n = [500(mmHG) × 30L] / [62.36(mmHGL/mol K) × 323.15K] = 0.744 mol

Moles of H₂O(v) is 0.1(0.744) = 0.0744 mol

Moles of N₂ is 0.9(0.744) = 0.6696 mol

we have that the moles of water condensed is 0.0744 mol i.e the water vapor  in the flask is condensed

Vн₂o = 0.0744 × 18 / 1 (g/cm³)

Vн₂o = 1.3392 cm³

Therefore, the  volume of the liquid water is 1.3392 cm³

(3). (ii) and (iii)

The absolute pressure of the gas and The partial pressure of water in the gas would change if the barometric pressure drops.

cheers i hope this helps!!!!

4 0
3 years ago
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