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sergeinik [125]
3 years ago
6

How many milliliters of C5H8 can be made from 366 mL C5H12 ?

Chemistry
1 answer:
zlopas [31]3 years ago
8 0

The number of Ml  of C₅H₈  that can  be  made  from 366  ml  C₅H₁₂  is 314.7 ml  of C₅H₈


  <u><em>calculation</em></u>

 step  1: write  the  equation for  formation of C₅H₈

C₅H₁₂  →  C₅H₈  + 2 H₂

Step 2: find the mass of C₅H₁₂

mass = density × volume

= 0.620 g/ml × 366 ml =226.92 g

Step 3: find moles  Of  C₅H₁₂

moles  = mass÷  molar mass

from periodic table the  molar mass of  C₅H₁₂ = (12 x5) +(  1 x12) = 72 g/mol

moles = 226.92 g÷ 72 g/mol =3.152 moles

Step 4: use the  mole ratio  to determine the  moles of C₅H₈

C₅H₁₂:C₅H₈  is 1:1  from equation above

Therefore the  moles of C₅H₈  is also = 3.152  moles

Step 5: find the mass  of C₅H₈

mass = moles x molar mass

from periodic table the  molar mass of C₅H₈ = (12 x5) +( 1 x8) = 68 g/mol

= 3.152  moles x 68 g/mol = 214.34 g

Step 6: find Ml of  C₅H₈

=mass / density

= 214.34 g/0.681 g/ml = 314.7 ml



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docker41 [41]

Answer:

pH = 10

The solution is basic.

Explanation:

A solution contains 1 × 10⁻⁴ M OH⁻ ions. First, we will calculate the pOH.

pOH = -log [OH⁻]

pOH = -log 1 × 10⁻⁴

pOH = 4

We can find the pH of the solution using the following expression.

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 4 = 10

Since the pH > 7, the solution is basic.

5 0
4 years ago
How many grams of iron are needed to produce 75.0 g of iron(III) chloride?
OleMash [197]

Answer:

Explanation:

Hello!

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In such a way, given the volumes and molarities of each reactant, we can compute the moles of produced iron (III) hydroxide by each of them, via the 3:1 and 1:1 mole ratios:

It means that the sodium hydroxide is the limiting reactant and 0.00833 moles of iron (III) hydroxide are produced; thus, the required mass is:

8 0
3 years ago
How many atoms are in 57.5 liters of xenon gas at STP
TiliK225 [7]

Answer:

1.55 × 10²⁴ atoms Xe

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 57.5 L Xe at STP

[Solve] atoms Xe

<u>Step 2: Identify Conversions</u>

[STP] 22.4 L = 1 mol

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 57.5 \ L \ Xe(\frac{1 \ mol \ Xe}{22.4 \ L \ Xe})(\frac{6.022 \cdot 10^{23} \ atoms \ Xe}{1 \ mol \ Xe})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 1.54583 \cdot 10^{24} \ atoms \ Xe

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe

4 0
3 years ago
3 H2 (g) + N2 (g) 2 NH3 (g)
ale4655 [162]

Answer:

Mass = 0.697 g

Explanation:

Given data:

Volume of hydrogen = 1.36 L

Mass of ammonia produced = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Solution:

Chemical equation:

3H₂ + N₂       →      2NH₃

First of all we will calculate the number of moles of hydrogen:

PV  = nRT

R = general gas constant = 0.0821 atm.L/mol.K

1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K

1.36 atm.L = n × 22.43 atm.L/mol

n = 1.36 atm.L / 22.43 atm.L/mol

n = 0.061 mol

Now we will compare the moles of hydrogen and ammonia:

                 H₂         :          NH₃

                  3          :            2

                0.061     :         2/3×0.061 = 0.041

Mass of ammonia:

Mass = number of moles × molar mass

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4 0
3 years ago
Metal rings can be coated with a layer of copper using electricity.
Eduardwww [97]

<u>First of all, what is electrolysis?</u>

Electrolysis is the process of breaking down ionic substances using direct current.

<u>Important points about electrolysis </u>

→ Ionic substances contain particles called ions.

→ Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions. The ions must be free to move for electrolysis to occur and it can happen by melting or dissolving an ionic substance in water.

→ Positively charged ions move to the negative electrode. They receive electrons and are <em>reduced</em>. The positive ions move towards the negative electrode because they want to cancel each other out.

→ Negatively charged ions move to the positive electrode.  They lose electrons and are <em>oxidised</em>. The substance that is broken down is called the electrolyte <em>(an electrolyte is just a liquid or solution that can conduct electricity)</em> . The negative ions move towards the positive electrode because they want to cancel each other out.

<h3>Cathode = Negative electrode</h3><h3>Anode = Positive electrode</h3>

Metal ions form at the cathode and non-metal ions form at the anode

How I remember if an element is <em>oxidised</em> or <em>reduced</em> is by remembering OIL RIG

OIL = Oxidation is Loss (of electrons)

RIG = Reduction is Gain (of electrons)

<h2><em><u>The answer to your question</u></em></h2>

1) The first step would be to clean the metal ring and sand it down because when the metal atoms from the electrolyte are deposited onto the ring, they will form a weak bond and they may simply 'fall' off. Also this could affect conductivity and the whole experiment. The more things you do accurately now, the more accurate your result will be.

2) You want to put the solution you are given in to the tank your going to be using.

3) This is basically the main part, you want to set up the circuit, I have attached a diagram at the bottom to show you the circuit. The copper rod will be the anode and the metal ring will be a cathode (ignore the elements).

4) Now turn on the circuit and you will start to see the solution spilt with the the solution now being split some going to the anode and some going the cathode.

5) Then a thin layer should form on the electrode.

Hope this helps :)

<h2><em><u></u></em></h2>

<em><u></u></em>

5 0
3 years ago
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