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2 months ago

93.2 mL of a 2.03 M potassium fluoride (KF) solution

2 answers:

6 0

Hello, in this situation, the initial volume was 93.2 mL, which converts to L as follows. After adding 3.92 L of water, the total volume of the final solution will encompass the water volume added, as illustrated here. Consequently, the molarity of the diluted potassium fluoride solution will be as follows. Best regards.

alisha [2.9K]2 months ago

3 0

Response:

1.98 M

Clarification:

Provided data

Starting volume (V₁): 93.2 mL

Starting concentration (C₁): 2.03 M

Water volume added: 3.92 L

Step 1: Convert V₁ to liters

Using the relationship 1 L = 1000 mL.

$93.2mL \times \frac{1L}{1000mL} = 0.0932 L$

Step 2: Calculate the final volume (V₂)

The final volume is the total of the initial volume and the added water volume.

$V_2 = 0.0932L + 3.92 L = 4.01L$

Step 3: Calculate the final concentration (C₂)

Utilizing the dilution rule.

$C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M$

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A 50.0 mL Volumetric flask filled with aqueous solution of hydrogen peroxide has a total mass of 88.5 g. The mass of the volumet

alisha [2963]

Answer:

1.44 g/mL

Explanation:

The following information was derived from the provided question:

Volume (V) of H2O2 = 50 mL.

Total mass of flask and H2O2 = 88.5 g.

Mass of the flask = 16.5 g.

What is the density (D)?

Next, we will calculate the mass of hydrogen peroxide (H2O2).

This is determined as follows:

Total mass of flask and H2O2 = 88.5 g.

Mass of the flask = 16.5 g.

What is the mass of H2O2?

Mass of H2O2 = (Total mass of flask + H2O2) – (Mass of flask)

Mass of H2O2 = 88.5 – 16.5

Mass of H2O2 = 72 g

Finally, we will find the density of hydrogen peroxide (H2O2) as follows:

Volume (V) of H2O2 = 50 mL.

Mass (m) of H2O2 = 72 g.

What is the density (D)?

Density (D) = mass (m) divided by volume (V)

D = m/V

D = 72 g / 50 mL

D = 1.44 g/mL

In conclusion, the hydrogen peroxide (H2O2) density is 1.44 g/mL.

6 0

2 months ago

Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculat

Tems11 [2777]

The question is not fully stated; here is the full version:

Using this data alongside the standard enthalpies of formation for $O_2(g)$ , $CO_2(g)$ , and $H_2O(l)$ found in Appendix C, determine the standard enthalpy of formation for acetone.

The complete combustion of 1 mole of acetone $(C_3H_6O)$ releases 1790 kJ:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l);\Delta H^o=-1790kJ$

Answer: The standard enthalpy of formation for $CO_2(g)$ is calculated to be -247.9 kJ/mol

Explanation:

Enthalpy change represents the variation in enthalpy for all products and reactants based on their respective mole counts. This is denoted as $\Delta H^o$

The enthalpy change calculation for a chemical reaction follows this equation:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

Concerning the chemical reaction in question:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)$

The equation reflecting the enthalpy change for this reaction is:

$\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_3H_6O(l))})+(4\times \Delta H^o_f_{(O_2(g))})]$

Provided data includes:

$\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-1790kJ$

Substituting values from the equation gives us:

$-1790=[(3\times {(-393.5)})+(3\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_6O(g))})+(4\times (0))]\\\\\Delta H^o_f_{(C_3H_6O(g))}=-247.9kJ/mol$

Thus, the enthalpy of formation of $C_3H_6O(g)$ computes to -247.9 kJ/mol.

4 0

2 months ago

HIPVs can cause what two more severe illnesses?

VMariaS [2998]

One of the conditions is Cancer.

4 0

2 months ago

A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells.

Alekssandra [3086]

Answer:

The right answer is "1.0100".

Explanation:

Assuming the total volume of the mixture is 100 ml.

Thus,

The volume of DMSO will be 10 mL and the volume of water will be 90 mL.

For DMSO:

= $10\times 1.1004$

= $11.004 \ g$

The total mass of the mixture will be:

= $90+11.004$

= $101.004 \ g$

Calculating the density of the mixture:

= $\frac{Mass}{Volume}$

= $\frac{101.004}{100}$

= $1.01004 \ g/mL$

Thus,

The specific gravity of the mixture is:

= $\frac{Density \ of \ mixture}{Density \ of \ water}$

= $\frac{1.01004}{1}$

= $1.0100$

3 0

2 months ago

What percent, by mass, is Oxygen in the compound Fe(OH)3?

Anarel [2989]

Answer:

Oxygen's mass percent in Fe(OH)3 is 44.92%

Explanation: The mass percentage is a means of indicating the concentration of a specific element within a compound. It is determined through the ratio of the element's mass to the compound's total mass, multiplied by 100.

•First calculate the overall mass of the compound

•Fe's molar mass = 55.85 g/mol

•O's molar mass = 16 g/mol

•H's molar mass = 1 g/mol

Using these values, we can compute the molecular mass of Fe(OH)3 = 55.85 g/mol + (16 g/mol)3 + (1 g/mol)3

=55.85 g/mol + 48 g/mol + 3 g/mol

=106.85 g/mol

Mass percent of an element = mass of element/total mass of compound × 100

In the case of 3 oxygen atoms present within the compound, the mass of oxygen totals 48 g/mol

Mass percent of oxygen= 48 g/mol/106.85 g/mol × 100

= 0.4492×100= 44.92%

[[TAG_31]]Thus, the mass percent of oxygen in Fe(OH)3 amounts to 44.92%[[TAG_32]]

7 0

2 months ago