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3 years ago

How many moles are in 19.6 g of Sodium (Na)? And the conversion factor?

Chemistry

1 answer:

Snezhnost [94]3 years ago

6 0

Answer:

n=0.852 moles

Explanation:

Given mass is, m = 19.6 g

The molar mass of sodium is, M = 22.99 u

We need to find the no of moles in 19.6 of Sodium. We know that, no of moles is equal to given mass divided by molar mass.

$n=\dfrac{m}{M}\\\\n=\dfrac{19.6}{22.99}\\\\n=0.852$

So, there are 0.852 moles in 19.6 g of Sodium.

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3 years ago

Read 2 more answers

0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium

Free_Kalibri [48]

Moles of potassium permanganate = 0.0008

<h3>Further explanation </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002$

From equation, mol ratio Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

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3 years ago

DUE TOMORROW!!! 15 POINTS

Lunna [17]

Answer:

A. write balanced chemical equation (including states), for this process.

Explanation:

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.

Hydrocarbon combustions always involve

[some hydrocarbon] + oxygen --> carbon dioxide + steam.

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side:

C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side:

C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

Now, we have fifteen oxygens on the right and O2 on the left.

Two ways to deal with that. We can use a fraction:

C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

Or, if you prefer to have whole number coefficients, double everything

to get rid of the fraction:

2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in...

C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

6 0

3 years ago

Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. kJ during a certain chemical reaction, w

puteri [66]

Answer:

= -356KJ

<em>therefore, the reaction where heat is released is exothermic reaction since theΔH is negative</em>

Explanation:

given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ

then,

change in enthalpy (ΔH) = -162( released energy)

work(w) = -194KJ

change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.

At constant pressure the change in enthalpy is given as

ΔH = ΔU + PΔV

ΔU = change in energy

ΔV = change in volume

P = pressure

w = -pΔV

therefore,

ΔH = ΔU -W

to evaluate energy change we have,

ΔU =ΔH + W

ΔU = -162+ (-194KJ)

= -356KJ

<em>therefore, the reaction where heat is released is exothermic reaction since theΔH is negative</em>

6 0

3 years ago