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2 years ago

What is the [h3o + ] in a 0.050 m solution of ba(oh)2?

2 answers:

7 0

The H3O+ in a 0.050M solution of Ba(OH)2 is calculated as below

write the equation for the dissociation of Ba(OH)2

Ba(OH)2 = Ba^2+ +2OH^-

calculate the OH- concentration

by use of mole ratio between Ba(OH)2 to OH^- which is 1:2 the concentration of OH = 0.050 x2 = 0.1 M

by use of the formula ( H3O+)(OH-) = 1 x10 ^-14

by making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-

substitute for OH-

H3O+ = (1 x10^-14 )/0.1

= 1 x10^-3 M

Ede4ka [16]2 years ago

4 0

Answer is: concentration of hydronium ions are 10⁻¹³ M.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.050 M.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
c(OH⁻) = 0.10 M = 10⁻¹ M.
c(OH⁻) · c(H₃O⁺) = 1·10⁻¹⁴ M².
c(H₃O⁺) = 10⁻¹⁴ M² ÷ 10⁻¹ M.
c(H₃O⁺) = 10⁻¹³ M.

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At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of

swat32

Ideal solutions obey Raoult's law, which states that:

P_i = x_i*(P_pure)_i

where
P_i is the partial pressure of component i above a solution
x_i is the mole fraction of component i in the solution
(P_pure)_i is the vapor pressure of pure component i

In this case,

P_benzene = 0.59 * 745 torr = 439.6 torr
P_toluene = (1-0.59) * 290 torr = 118.9 torr

The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:

P_total = (439.6 + 118.9) torr = 558.5 torr

Assuming the gas phase also behaves ideally, the partial pressure of each gas in the vapor phase is proportional to its molar concentration, so the mole fraction of toluene in the vapor phase is:

118.9 torr/558.5 torr = 0.213

8 0

2 years ago

What is the percent yield of sodium hydroxide in the given reaction? The reaction was performed using 45 g NaHCO3 and 18 g NaOH

Vlada [557]

Answer:

The percent yield = 83.33 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $NaHCO_3$ :-

Mass of water = 45 g

Molar mass of $NaHCO_3$ = 84.007 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{45\ g}{84.007\ g/mol}$

$Moles\ of\ NaHCO_3= 0.54\ mol$

For $NaOH$ :-

Given mass = 18 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{18\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.45\ mol$

According to the given reaction:

$NaHCO_3\rightarrow NaOH+CO_2$

1 mole of $NaHCO_3$ on reaction forms 1 mole of $NaOH$

Thus,

0.54 mole of $NaHCO_3$ on reaction forms 0.54 mole of $NaOH$

Moles of $NaOH$ formed = 0.54 moles

Moles of $NaOH$ actually formed = 0.45 moles

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 0.54 moles

Experimental yield = 0.45 moles

Applying the values in the above expression as:-

$\%\ yield =\frac{0.45}{0.54}\times 100$

$\%\ yield =83.33\ \%$

The percent yield = 83.33 %

7 0

1 year ago

A mixture of ethyl alcohol (molecular mass = 46.1 u) and water (molecular mass = 18.0 u) contains one mole of molecules. The mix

Ne4ueva [31]

Answer:

10.2 g

Explanation:

A mixture of ethyl alcohol (molecular mass = 46.1 u) and water (molecular mass = 18.0 u) contains one mole of molecules.

The mixture contains 20.0 g of ethyl alcohol. The molar mass of ethyl alcohol is 46.1 g/mol. The moles of ethyl alcohol are:

20.0 g × (1 mol / 46.1 g) = 0.434 mol

The sum of the moles of ethyl alcohol and the moles of water is 1 mole.

n(ethyl alcohol) + n(water) = 1 mol

n(water) = 1 mol - n(ethyl alcohol) = 1 mol - 0.434 mol = 0.566 mol

The molar mass of water is 18.0 g/mol. The mass of water is:

0.566 mol × (18.0 g/mol) = 10.2 g

7 0

2 years ago

If 5.10 g of sodium and 305 g of potassium nitrate react in an airbag, how many grams of kno3 remain because of the limited amou

Oksi-84 [34.3K]

The balanced equation for the above reaction is as follows;
10Na(s) + 2KNO₃(s)--> K₂O(s) + 5Na₂O(s) + N₂(g)
Stoichiometry of Na to KNO₃ is 10 : 2
Number of Na moles reacted - 5.1 g/ 23 g/mol = 0.22 mol
Na is the limiting reactant, therefore Na is fully used up, Since KNO₃ is present in excess , at the end of the reaction a certain amount of KNO₃ will be remaining.
10 mol of Na reacts with 2 mol of KNO₃
Therefore 0.22 mol of Na reacts with - 2 /10 x 0.22 = 0.044 mol of KNO₃
Mass of KNO₃ reacted = 0.044 mol x 101.1 g/mol = 4.45 g
Mass of KNO₃ present initially - 305 g
Therefore remaining mass of KNO₃ - 305 - 4.45 = 300.55 g

7 0

2 years ago

Enough water is added to 50.0 mL of a 0.660 M NaOH solution in order to bring the total volume to 450.0 mL. What is the molarity

g100num [7]

Answer:

0.073 M

Explanation:

M1*V1 = M2*V2

M1 = 0.660 M

V1 = 50.0 mL

V2 = 450.0 mL

0.660M * 50.0 mL = M2*450.0 mL

M2 = 0.660M*50.0/450.0 = 0.073 M

5 0

2 years ago