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zubka84 [21]
3 years ago
9

What is the mass in grams of 7.5 mol of C8H18?

Chemistry
1 answer:
Lana71 [14]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 856.74 \ g \ C_8H _{18}}}

Explanation:

To convert from grams to moles, the molar mass is used. These values tells us the grams in 1 mole of a substance. They can be found on the Periodic Table (they are equivalent to the atomic masses, but the units are grams per mole).

We are given the compound C₈H₁₈. Look up the molar masses of the individual elements.

  • Carbon (C): 12.011 g/mol
  • Hydrogen (H): 1.008 g/mol

Notice there are subscripts that tell us the number of atoms of each element. We must multiply the molar masses by the subscripts.

  • C₈: 8(12.011 g/mol)=96.088 g/mol
  • H₁₈: 18(1.008 g/mol)=18.144 g/mol

Add these 2 values together to find the molar mass of the whole compound.

  • C₈H₁₈: 96.088 g/mol +18.144 g/mol=114.232 g/mol

Use this number as a ratio.

\frac{ 114.232 \ g \ C_8H_{18}}{1 \ mol \ C_8H_{18}}

Multiply by the given number of moles: 7.5

7.5 \ mol \ C_8H_{18}*\frac{ 114.232 \ g \ C_8H_{18}}{1 \ mol \ C_8H_{18}}

The moles of C₈H₁₈ will cancel each other out.

7.5 *\frac{ 114.232 \ g \ C_8H_{18}}{1}

7.5 *{ 114.232 \ g \ C_8H_{18}}

856.74 \ g \ C_8H _{18}

7.5 moles of C₈H₁₈ is equal to <u>856.74 grams of C₈H₁₈</u>

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There are  20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.

<h3>How many molecules in 3.4 moles of NH4NO3?</h3>

We know that one mole of a substance has  6.022 × 10²³ molecules so in 3.4 moles of NH4NO3, we have 20.5 x 10^24 molecules if we multiply the  6.022 × 10²³ with 3.4.

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Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100
yulyashka [42]

Answer:

ΔH = -59.6kJ/mol

Explanation:

The reaction that occurs between Ag⁺ and Cl⁻ ions is:

Ag⁺ + Cl⁻ → AgCl(s) + ΔH

To find ΔH we need to obtain moles of reaction and heat released in the reaction because ΔH is defined as heat released per mole of reaction.

<em>Moles of reaction:</em>

Moles of Ag⁺ and Cl⁻ added are:

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Cl⁻: 0.100L * (0.200mol / L) 0 0.02 moles

That means limiting reactant is Ag⁺ and moles of reaction are 0.01 moles

<em>Heat released:</em>

To find heat released we must use coffe cup calorimeter equation:

Q = C*m*ΔT

<em>Where C is specific heat of solution (4.18J/g°C), m is the mass of solution (200g because there are 100 + 100mL = 200mL and density of solution is 1g/mL) and ΔT is change in temperature (25.30°C - 24.60°C = 0.70°C).</em>

Replacing:

Q = C*m*ΔT

Q = 4.18J/g°C * 200g * 0.70°C

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Is total heat released.

The calorimeter absorbs:

15.5J / °C * 0.7°C = 10.85

Thus, when 0.01 moles reacts, 585.2J + 10.85  = 596.05J are released (Heat released is heat abosrbed by calorimeter + Heat absorbed by water) and ΔH is:

ΔH = 596.05J / 0.01 moles =

ΔH = 59605J / mol =

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<em>As heat is released, ΔH < 0.</em>

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