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never [62]
2 years ago
15

Calculate the number of grams of xenon in 4.658 g of the compound xenon tetrafluoride.

Chemistry
1 answer:
andrezito [222]2 years ago
5 0

Answer:

The mass of xenon in the compound is 2.950 grams

Explanation:

Step 1: Data given

Mass of XeF4 = 4.658 grams

Molar mass of XeF4 = 207.28 g/mol

Step 2: Calculate moles of XeF4

Moles XeF4 = mass XeF4 / molar mass XeF4

Moles XeF4 = 4.658 grams / 207.28 g/mol

Moles XeF4 = 0.02247 moles

Step 3: Calculate moles of xenon

XeF4 → Xe + 4F-

For 1 mol xenon tetrafluoride, we have 1 mol of xenon

For 0.02247 moles XeF4 we have 0.02247 moles Xe

Step 4: Calculate mass of xenon

Mass xenon = moles xenon * molar mass xenon

Mass xenon = 0.02247 moles * 131.29 g/mol

Mass xenon = 2.950 grams

The mass of xenon in the compound is 2.950 grams

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On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
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Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

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Which statement accurately compares ionic and covalent bonding?
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Answer:

Conduct electricity when they are molten, while covalent compounds usually do not conduct electricity when they are molten.

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Based on the results of this lab, write a short paragraph that summarizes how to distinguish physical changes from chemical chan
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A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition, the student measures out 5.8
ra1l [238]

Answer:

<u>1. Net ionic equation:</u>

  • Cl⁻(aq) +  Ag⁺(aq) → AgCl(s)

<u />

<u>2. Volume of 1.0M AgNO₃</u>

  • 41ml

Explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

  • (aq) for ions in aqueous solution
  • (s) for solid compounds or elements
  • (g) for gaseous compounds or elements

  • NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

  • NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
  • AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
  • NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

  • Na⁺(aq) + Cl⁻(aq) +  Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) +  Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

  • Cl⁻(aq) +  Ag⁺(aq) → AgCl(s) ← answer

2.  How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

  • number of moles = mass in grams / molar mass
  • molar mass of AgCl = 143.32g/mol
  • number of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

  • molarity = number of moles of solute / volume of solution in liters

  • 1.0M = 0.040748mol / V

  • V = 0.040748mol / (1.0M) = 0.040748 liter

  • V = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer

4 0
2 years ago
120 grams of calcium nitrite Ca(NO2)2 is dissolved in a 240 mL solution. What is the molarity of the solution???
expeople1 [14]

Hey there!:

Molar mass Ca(NO2)2 = 132.089 g/mol

Mass of solute = 120 g

Number of moles:

n = mass of solute / molar mass

n = 120 / 132.089

n = 0.0009084 moles of Ca(NO2)2

Volume in liters of solution :

240 mL / 1000 => 0.24 L

Therefore:

Molarity = number of moles / volume of solution

Molarity = 0.0009084 / 0.24

Molarity = 0.003785 M

Hope that helps!

5 0
2 years ago
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