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sammy [17]
1 year ago
7

Calculate the molarity of each solution.

Chemistry
1 answer:
weeeeeb [17]1 year ago
8 0
Q1)
molarity is defined as the number of moles of solute in 1 L solution 
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L 
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol 
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol 
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL 
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M
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How many equivalents are in 0.60 mole of Mg2+
Allisa [31]

Equivalents express the combining power of an element for example what mass of an element can combine with or displace one gram of hydrogen or eight grams of oxygen.

Since Mg is divalent, one gram of Mg2+ can combine with 2 grams of hydrogen  

Therefore, 0.6 mole of Mg2+ can combine with 0.6 x 2 = 1.2 moles of hydrogen

Hence there are 1.2 equivalents in 0.60 mole of Mg2+

5 0
2 years ago
A sample of hydrogen gas was collected over water at 21ºc and 685 mmhg. the volume of the container was 7.80 l. calculate the m
murzikaleks [220]
To determine the mass of the hydrogen gas that was collected, we calculate for the moles of hydrogen gas from the conditions given. In order to do this, we need an equation which would relate pressure, volume and temperature. For simplicity, we assume the gas is an ideal gas so we use the equation PV = nRT where P is the pressure, V is the volume, n is the number of moles of the gas, T is the temperature and R is the universal gas constant. We calculate as follows:

PV = nRT
n = PV / RT
n = (18.6/760) (7.80) / 0.08205 ( 21 + 273.15)
n = 0.0079 mol

Mass = 0.0079 mol ( 18.02 g / mol ) = 0.1425 g H2
8 0
2 years ago
B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 reacts
n200080 [17]

Answer:

Heat realesed is - 2192.7 kJ

Explanation:

First lets have the chemical equation balanced, and then solve the question based on the fact that the enthalpy change for a reaction is the sum of the enthalpies of formation of products minus reactants.

    B₅H₉ (l) +     O₂ (g) ⇒       B₂O₃ (s) +      H₂O(l)

B atoms we have 5 reactants and 2 products, so the common mltiple is 10 and we have

  2  B₅H₉ (l) +     O₂ (g) ⇒      5 B₂O₃ (s) +      H₂O(l)

Now balance H  by multiplying by 9 the H₂O

  2  B₅H₉ (l) +     O₂ (g) ⇒      5 B₂O₃ (s) +    9 H₂O(l)

Finally, balance the O since we have 24 in products by multiplying by 12 the

O₂ ,

        2  B₅H₉ (l) +  12 O₂ (g) ⇒      5 B₂O₃ (s) +    9 H₂O

ΔHrxn = 5 x ΔHºf B₂O₃  + 9 x   ΔHºf H₂O  -  ( 2 x ΔHºf  B₅H₉ + 12 x  ΔHºf O₂  )

We have all the  ΔHºf s except oxygen but remember the enthalpy of formation of a pure element in its standard estate is cero.

ΔHrxn =  5 mol x ( -1272 kJ/mol )+ 9 mol  x ( -285.4 kJ/mol )   - (  2mol x 732 kJ/mol )

= -8928.6 kJ - 1464  kJ = -10,392 kJ

Now this enthalpy change was based in 2 mol  reacted according to the balanced equation, so for 0.211 mol of  B₅H₉  we will have:

-10,392 kJ/ 2 mol B₅H₉   x 0.211 mol B₅H₉ = - 2192.7 kJ

8 0
1 year ago
Read 2 more answers
What mass of oxygen reacts when 84.9 g of iron is consumed in the following reaction: Fe+O2= Fe2O3
konstantin123 [22]
First we will calculate the number of moles of Iron:
n =  \frac{m}{M}
, where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
n= \frac{84.9}{55.845} = 1,52 moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So 1,52= \frac{mOxygen}{32} where 32 is the Atomical Weight of Oxygen (16 x 2).
=>mOxygen=32*1,52=48,64g
6 0
1 year ago
Read 2 more answers
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0
2 years ago
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