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Anestetic [448]
3 years ago
11

What is the volume in liters of 3.51kg of ethylene glycol and density of 1.11g/cm^3?

Chemistry
1 answer:
AVprozaik [17]3 years ago
4 0
D = m/v ---> v = m/d

v = 3162 cm^3
   = <span>0.003162000m^3

</span>
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Neon is an element and elements are pure oxygen is. Or it is a gas
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3 years ago
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Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio
soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂  → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • CuCl₂: 1 mole
  • CuCO₃: 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

  • Na₂CO₃: 129 g/mole
  • CuCl₂: 134.45 g/mole
  • CuCO₃: 123.55 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Na₂CO₃: 1 mole ×129 g/mole= 129 grams
  • CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
  • CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
  • NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

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6 0
1 year ago
A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The t
svp [43]
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225 
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>

<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol


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Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the
almond37 [142]

The powder could be acetaminophen, analgesic  having chemical formula C_{17}H_{21}NO_{4}

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

Percentage of H = 6.978 %

Percentage of N = 4.617 %

Percentage of O = 21.10 %

Let the mass of the compound be 100 g. So, the percentages given are taken as mass.

Mass of C = 67.31 g

Mass of H = 6.978 g

Mass of N = 4.617 g

Mass of O = 21.10 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of carbon = \frac{mass}{molar \;mass}

Moles of carbon =\frac{67.31g}{12g/mole}

=5.60 moles

Moles of hydrogen = \frac{mass}{molar \;mass}

Moles of hydrogen = \frac{6.978 g}{1 g/mole}

=6.978 moles

Moles of nitrogen =\frac{mass}{molar \;mass}

Moles of nitrogen = \frac{4.617  g}{14 g/mole}

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Moles of oxygen =\frac{mass}{molar \;mass}

Moles of oxygen =\frac{21.10 g}{16 g/mole}

=1.31 moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is C_{17}H_{21}NO_{4}.

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3 0
2 years ago
If a substance has a half life of 45 days, how much of the original substance would be left after 90 days?
Fofino [41]
U would subtract 90 minus 45 which equals 45
8 0
3 years ago
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