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What is the number of moles of Sodium Chloride (NaCl) that is present in 29.22g sample of it?

Liula [17]

There are 0.5 moles of NaCl present in a 29.22g sample.

to find moles of a compound you would look at the periodic table to find the mass. in this case, the mass of NaCl is 58.44 g/mol. now, you divide the sample mass of 29.22g by 58.44 g/mol. the grams will cancel each other out and you’re left with 0.5 mol in that sample

hope this helps!

3 0

3 years ago

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Use the reaction and bond information to answer the question. C2H6 → C2H4 H2 Reactant bond energies: H–C = 413 kJ/mol, C–C singl

nlexa [21]

The energy of the reactant has been higher than the energy of the product, Thus, the reaction has been an exothermic reaction.

Bond energy has been defined as the chemical energy possessed by the bonds in the chemical structure. The energy has been conserved in the chemical reaction.

<h3>Is the reaction endothermic or exothermic?</h3>

In the reaction in which bond energy of the reactant has been higher than the product, it has been an exothermic reaction.

In the reaction in which the bond energy of the product is higher than the reactant, it has been an endothermic reaction.

In the given reaction

$\rm C_2H_6\;\rightarrow\;C_2H_4\;+\;H_2$

The bond energies of reactant has been:

H-C bond = 413 kJ/mol
C-C bond = 347 kJ/mol

The total bond energy of the reactant has been:

$\rm Reactant=6\;\times\;H-C\;bond\;+\;C-C\;bond \\Reactant=6\;\times\;413+347\;kJ/mol\\Reactant=2,825\;kJ/mol$

The total bond energy of the reactant has been 2,825 kJ/mol.

The bond energies of product has been:

H-C = 413 kJ/mol
C=C = 614 kJ/mol
H-H = 432 kJ/mol

The total bond energy of the product has been:

$\rm Product=4\;\times\;H-C\;bond\;+\;C=C\;bond\;+\;H-H\;bond\\Product=4\;\times\;413\;+\;614\;+\;432\;kJ/mol\\Product=2,698\;kJ/mol$

The total bond energy of the product has been 2,698 kJ/mol.

The energy of the reactant has been higher than the energy of the product, Thus, the reaction has been an exothermic reaction, as energy can neither be created nor be destroyed. It has been conserved in the chemical reaction.

Learn more about exothermic reaction, here:

brainly.com/question/26203526

4 0

3 years ago

What volume of 1.00 m hcl in liters is needed to react completely (with nothing left over) with 0.750 l of 0.100 m na2co3?

kotykmax [81]

The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L

3 0

3 years ago

Please help me answer this I will be very grateful if so :)

dimaraw [331]

The molecules in system #2 have a higher kinetic energy because they are at a higher temperature than molecules in system#1.

<h3>Heating of water molecules</h3>

Temperature is defined as a measure of the average kinetic energy of the molecules of a body. The higher the temperature of a body, the higher the kinetic energy of the molecules of the body.

In both systems, we have water molecules that have the same formula H2O. However, the molecules in system #2 have a higher kinetic energy because they are at a higher temperature than molecules in system#1.

Learn more about kinetic energy of molecules: brainly.com/question/2731193

8 0

3 years ago

To determine the concentration of 20.00 mL of an unknown solution of a monoprotic acid, it is titrated with 0.1093 M sodium hydr

Elanso [62]

Answer:

d) 0.1202 M

Explanation:

Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.

NaOH + HA → NaA + H₂O

The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:

22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol

The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.

The molar concentration of HA is:

2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M

7 0

3 years ago