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3 years ago

A 10.0 mL syringe contains 0.10 g of nitrogen gas at 0.0 degrees Celsius. What is the pressure inside the syringe?

1 answer:

4 0

Since the sample inside the syringe is a gas, we assume that it is an ideal gas that is we use the expression PV = nRT. We can easily calculate the pressure as follows:

n = m /MM = 0.10 / 28 = 0.0036 mol

P = nRT / V = 0.0036(0.08206)(273.15) / 0.010
P = 8 atm

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You wish to prepare an HC2H3O2 buffer with a pH of 4.24. If the pKa of is 4.74, what ratio of C2H3O2 /HC2H3O2 must you use?

LenaWriter [7]

To solve this problem, we can use the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:

pH = pKa + log ([base]/[acid]) ---> 1

Where,

[base] = concentration of C2H3O2 in molarity or moles

[acid] = concentration of HC2H3O2 in molarity or moles

For the sake of easy calculation, let us assume that:

[base] = 1

[acid] = x

Therefore using equation 1,
4.24 = 4.74 + log (1 / x)

log (1 / x) = - 0.5

1 / x = 0.6065

x = 1.65

The required ratio of C2H3O2 /HC2H3O2 is 1:1.65 or 3:5.

3 0

3 years ago

A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If a 100.-mL sample of the HCl solution

madam [21]

Answer: The initial pH of the HCl solution is 3

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=100mL\\n_2=1\\M_2=0.10M\\V_2=1mL$

Putting values in above equation, we get:

$1\times M_1\times 100=1\times 0.10\times 1\\\\M_1=\frac{1\times 0.10\times 1}{1\times 100}=10M$

1 mole of HCl produces 1 mole of $H^+$ ions and 1 mole of $Cl^-$ ions

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=0.001M$

Putting values in above equation, we get:

$pH=-\log (0.001)\\\\pH=3$

Hence, the initial pH of the HCl solution is 3

6 0

3 years ago

Having a child will change your financial obligations, and can lead to a career change.

MrRa [10]

Yes, that's a true statement?

3 0

3 years ago

Read 2 more answers

A solution is prepared by mixing 200.0 g of water, H2O, and 300.0 g of

VikaD [51]

Answer:

Mole Fraction (H₂O) = 0.6303

Mole Fraction (C₂H₅OH) = 0.3697

Explanation:

(Step 1)

Calculate the mole value of each substance using their molar masses.

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g

Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₂H₅OH): 46.068 g/mol

300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g

(Step 2)

Using the mole fraction ratio, calculate the mole fraction of each substance.

moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent

11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (H₂O) = 0.6303

6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (C₂H₅OH) = 0.3697

7 0

1 year ago

Write a balanced equation for the reaction of Na₂S and CdSO₄. Express your answer as a chemical equation. Identify all of the ph

Luba_88 [7]

Explanation:

A chemical reaction equation that contains same number of atoms on both reactant and product side is known as a balanced chemical reaction equation.

For example, $Na_{2}S(aq) + CdSO_{4}(aq) \rightarrow Na_{2}SO_{4}(aq) + CdS(s)$

Here, number of reactant molecules are as follows.

Na = 2

S = 2

Cd = 1

O = 4

Number of product molecules are as follows.

Na = 2

S = 2

Cd = 1

O = 4

As there are already equal number of atoms on both reactant and product side. Hence, the reaction equation is balanced.

Also, in this reaction both $Na_{2}S$ and $CdSO_{4}$ are present as aqueous solution. And, CdS is the precipitate that is formed as it is the insoluble solid that forms.

$Na_{2}SO_{4}$ is present in aqueous state.

3 0

3 years ago