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14 days ago

9

34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfu

ric acid solution?

1 answer:

Anarel [2.5K]14 days ago

5 0

0.1045M represents the concentration of the initial sulfuric acid solution

Explanation:

Titration is conducted to ascertain the volume or concentration of an unknown electrolyte.

Given data:

acid volume = 50 ml

acid molarity =?

NaOH volume = 34.62 ml

molarity of NaOH = 0.1510

The titration formula applied is

Macid x Vacid = Mbase x Vbase

Substituting the values into the above equation results in:

Macid x 50 = 34.62 x 0.1510

Macid = $\frac{5.22762}{50}$

= 0.1045 M, indicating the molarity of the sulfuric acid solution used to neutralize the 0.1510 M base solution.

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1) Calcium carbonate comprises 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (which is 40.0%)!

2) The mass fraction mentioned is superfluous information.

3) The resulting solution is:

m(Ca)=1.2 g

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m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g

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29 days ago

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7.744 Liters of nitrogen are contained in a container. Convert this amount to grams.

lorasvet [2510]

Answer:

9.69g

Explanation:

To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.

1 mole of gas takes up 22.4L at STP.

Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning

X moles of N2 = 7.744/22.4 = 0.346 moles

Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:

Molar Mass of N2 = 2x14 = 28g/mol

Number of moles N2 = 0.346 moles

Find the mass of N2 =?

Mass = number of moles × molar mass

Mass of N2 = 0.346 × 28

Mass of N2 = 9.69g

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1 month ago

A 30.0 mL sample of hydrogen gas (H2) is collected over water at 20.00∘C and has a total pressure of 700.0 torr. The partial pre

Alekssandra [2698]

Answer: The mole fraction of hydrogen gas at 20°C is 0.975

Explanation:

The information provided includes:

Water vapor pressure at 20°C is 17.5 torr

Total pressure at 20°C = 700.0 torr

Hydrogen gas vapor pressure at 20°C = (700.0 - 17.5) torr = 682.5 torr

To find hydrogen gas's mole fraction at 20°C, we utilize Raoult's law, represented by:

$p_{H_2}=p_T\times \chi_{H_2}$

where,

$p_{H_2}$ = pressure of hydrogen gas = 682.5 torr

$p_T$ = total pressure = 700.0 torr

$\chi_{H_2}$ = mole fraction of hydrogen gas =?

Substituting the values into the equation yields:

$682.5torr=700.0torr\times \chi_{H_2}\\\\\chi_{H_2}=\frac{682.5}{700.0}=0.975$

Thus, the mole fraction of hydrogen gas at 20°C equals 0.975

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20 days ago

PART A: Use the following glycolytic reaction to answer the question. If the concentration of DHAP is 0.125 M and the concentrat

alisha [2704]

Answer:

For A: The change in free energy for the reaction is -5339.76 J/mol

For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

Explanation:

Regarding the specified chemical reaction:

$DHAP\rightleftharpoons G_3P$

For A:

The relationship between standard Gibbs free energy and equilibrium constant is as follows:

$\Delta G^o=-RT\ln K_{eq}$

The free energy change can be calculated using the following equation:

$\Delta G=\Delta G^o+RT\ln Q$

Or,

$\Delta G=-RT^o\ln K_{eq}+RT\ln Q$

where,

$\Delta G$ = Change in free energy

R = Gas constant = $8.314J/K mol$

$T^o$ = standard temperature = $25^oC=[273+25]K=298K$

T = temperature of the cell = $37^oC=[273+37]K=310K$

$K_[eq}$ = equilibrium constant = $5.4\times 10^{-2}$

Q = reaction quotient = $\frac{[G_3P]}{[DHAP]}$

$[G_3P]$ = 0.06 M

[DHAP] = 0.125 M

Substituting the values into the equation yields:

$\Delta G=[-(8.314J/mol.K\times 298K\times \ln (5.4\times 10^{-2}))]+[(8.314J/mol.K\times 310K\times \ln (\frac{0.06}{0.125}))]\\\\\Delta G=-[-7231.46]+[-1891.7]=-5339.76J/mol$

Thus, the change in free energy for the reaction is -5339.76 J/mol

For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So, $-5339.76J/mol\times \frac{1kJ}{1000J}=-5.34kJ/mol$

Consequently, the free energy change's units are kJ/mol

For C:

For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

Consequently, the forward reaction is favored and the reverse reaction is not favored.

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27 days ago

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21 day ago

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