Is my answer right? How many millilitres of 1.33 mol L−1 H2SO4(aq) are required to completely neutralize 49.3 mL of 0.830 mol L−1 KOH(aq) ?

I got 15.4 once and now I got 61.4? Are they correct? if so, which one?

Answers

Answer 1

Answer:

Final answer:

To neutralize the KOH solution, we need 61.4 mL of 1.33 mol L−1 H2SO4(aq).

Explanation:

To find the volume of the H2SO4 solution needed to neutralize the KOH solution, we can use the equation:

Mole of H2SO4 = Molarity of KOH x Volume of KOH

First, calculate the moles of KOH:
Moles of KOH = Molarity of KOH x Volume of KOH = 0.830 mol/L x (49.3 mL / 1000 mL) = 0.04089 mol

Since H2SO4 is a diprotic acid and KOH is a strong base, the reaction will be:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O

Therefore, the ratio between the moles of H2SO4 and KOH is 1:2. This means that twice the moles of KOH will be needed to neutralize the H2SO4. Calculate the moles of H2SO4 needed:
Moles of H2SO4 needed = 2 x Moles of KOH

= 2 x 0.04089 mol

= 0.08178 mol

Finally, calculate the volume of the H2SO4 solution needed:
Volume of H2SO4 = Moles of H2SO4 / Molarity of H2SO4 = 0.08178 mol / 1.33 mol/L

= 0.0614 L

= 61.4 mL

Learn more about Neutralization here:

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A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution? Use the periodic table in the toolbar if needed.

Answers

Answer : The mass of NaOH present in the solution is, 0.0625 grams

Explanation : Given,

Mass % = 25 %

Mass of solution = 0.250 g

Formula used :

$Mass\%=\frac{\text{Mass of}NaOH}{\text{Mass of solution}}* 100$

Now put all the given values in this formula, we get the mass of NaOH.

$25=\frac{\text{Mass of}NaOH}{0.250g}* 100$

$\text{Mass of}NaOH=0.0625g$

Therefore, the mass of NaOH present in the solution is, 0.0625 grams

The concentration of NaOH is 25.00% by mass, it means that 25.00% of the mass of the solution is of NaOH. Hence:

$m_(NaOH)=25.00\%* m_(solution)\Longrightarrow m_(NaOH)=(25)/(100)* 0.250~g\iff\n\n\boxed{m_(NaOH)=0.0625~g}$

In a constantvolume bomb calorimeter, the combustion of 0.6654 gof an organic
compound with a molecular mass of 46.07 amu causesthe temperature
in the calorimeter to rise from 25.000oC to 30.589
oC. The total heat capacity ofthe calorimeter and all
its contents is 3576 JoC-1. What is
the energy of combustion ofthe organic compound,
DU/ kJ
mol-1?

Answers

Answer:

1383.34 kJ/mol is the energy released on combustion of the organic compound.

Explanation:

Mass of an organic compound = 0.6654 g

Molar mass of organic compound = 46.07 g/mol

Moles of an organic compound = $(0.6654 g)/(46.07 g/mol)=0.01444 mol$

Let heat evolved during burning of 0.6654 grams of an organic compound be -Q.

Heat absorbed by calorimeter = Q' = -Q

The total heat capacity of the calorimeter all its contents = C

C = 3576 J/°C

Change in temperature of the calorimeter =

ΔT = 30.589°C - 25.000°C = 5.589°C

$Q'=C* \Delta T$

$Q'=3576 J/^oC* 5.589^oC=19,975.536 J=19.975 kJ$

Q' = 19.975 kJ

Q = -19.975 kJ (negative sign; energy released)

0.01444 moles of an organic compound gives 19.975 kilo Joule.

The 1 mole of an organic compound will give : $\Delta H_(comb)$

$\Delta H_(comb)=(-19.975 kilo Joule)/(0.01444 mol)$

$=-1383.34 kJ/mol$

How could you use these solutions to determine the identities of each metal powder?Fill in the blanks with options in below:
1. The nitric acid solution will oxidize and thus dissolve _________. This will allow to identify ________.
2. To distinguish between ________, we can use the nickel nitrate
3. The nickel nitrate solution will oxidize and thus dissolve ________ and will not oxidize or dissolve ________.
Options:
a. Zn and Pt
b. Zn, Pb and Pt
c. Pb and Pt
d. Pb
e. Zn
f. Pt
g. Zn and Pb

Answers

Answer:

1.) The nitric acid solution will oxidize and thus dissolve _*(Zn and Pb)*_. This will allow to identify _**Pt**_.

2) To distinguish between _*(Zn and Pb)*_, we can use the nickel nitrate.

3) The nickel nitrate solution will oxidize and thus dissolve _**Zn**_ and will not oxidize or dissolve _**Pb**_.

Explanation:

1) Unlike Zinc and Lead, Platinum does not react with Nitric acid. So, it will be the only metal from step 1 that doesn't react. Pt is identified in this manner.

2) Nickel is higher than Lead in the activity series, but Zinc is higher than both of them in the activity series. This selectivity can be used to distinguish between Zinc and Lead metal powders.

3) Because Zinc is higher than Nickel in the activity series, it means that Zinc metal can and will displace Nickel from Nickel Nitrate solution. Therefore the Nickel Nitrate solution will oxidize and dissolve the Zinc metal.

But, there will be no reaction with the Lead metal powders sample as Pb is lower than Ni in the activity series, so, Nickel Nitrate solution will not oxidize or dissolve the Lead metal powders.

1. Potassium (K) has an atomic mass of 39.0983 amu and only two naturally-occurring isotopes. The K-41 isotope (40.9618 amu) has a natural abundance of 6.7302%. What is the mass (in amu) of the other isotope

Answers

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618 $*$ 0.067302) + ( ? $*$ (1-0.067302)

39.0983 = 2.756811064 + ( ? $*$ 0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ? $*$ 0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618 0.067302) + ( ? (1-0.067302)

39.0983 = 2.756811064 + ( ? 0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ? 0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

Systematic name for: Silver Nitrate.molar mass of Ag:
molar mass of N:
molar mass of O:

and the overall molar mass for Silver Nitrate.

Answers

Answer:

$AgNO_3$ ,Molar mass =169.87 g/mol

Explanation:

The systematic name for silver nitrate is $AgNO_3$

Now we have to calculate the molar mass of $AgNO_3$

Molar mass of Ag = 107.87 g/mol

Molar mass of N =14 g/mol

Molar mass of O =16 g/mol

So the molar mass of silver nitrate ( $AgNO_3$ ) is

$=(1* 107.87+1* 14+3* 16)=169.87g/mol$

Calculate the volume of 0.100 m hcl required to neutralize 1.00 g of ba(oh)2 (molar mass = 171.3 g/mol).

Answers

The balanced chemical equation between HCl and $Ba(OH)_(2)$ is:

$2HCl (aq) + Ba(OH)_(2)(aq) -->BaCl_(2)(aq) + 2 H_(2)O(l)$

Moles of $Ba(OH)_(2)$ = $1.00 g Ba(OH)_(2) * (1 mol Ba(OH)_(2))/(171.3 g Ba(OH)_(2)) = 0.00584 mol Ba(OH)_(2)$

Moles of HCl required to neutralize $Ba(OH)_(2)$ :

$0.00584 mol Ba(OH)_(2) * ( 2 mol HCl)/(1 mol Ba(OH)_(2)) = 0.01168 mol HCl$

Calculating the volume of HCl from moles and molarity:

$0.01168 mol HCl * (1 L)/(0.100 mol) * (1000 mL)/(1 L) = 116.8 mL$

Answer:- 117 mL of HCl are used.

Solution:- The balanced equation for the reaction of HCl with barium hydroxide is written as:

$2HCl(aq)+Ba(OH)_2(aq)\rightarrow BaCl_2(aq)+2H_2O(l)$

From above equation, HCl and $Ba(OH)_2$ react in 2:1 mol ratio.

We will calculate the moles of barium hydroxide on dividing its grams by its molar mass.

Molar mass of Barium hydroxide is given as 171.3 g per mol.

$1.00gBa(OH)_2((1mol)/(171.3g))$

= 0.00584 mol $Ba(OH)_2$

Using mol ratio we calculate the moles of HCl as:

$0.00584molBa(OH)_2((2molHCl)/(1molBa(OH)_2))$

= 0.01168 mol HCl

We know that molarity is moles of solute per liter of solution. We have 0.01168 moles of HCl and its molarity is 0.100 M. So, we can calculate the liters of HCl solution used on dividing the moles by molarity as and on multiplying by 1000 the liters are converted to mL since, 1 L = 1000 mL.

$0.01168mol((1L)/(0.100mol))((1000mL)/(1L))$