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3 years ago

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A solution of [FeSCN]^{2+} is found to have 37% transmittance at 447 nm. If the molar absorption coefficient (ε) is 4400 at λ_{m

ax} 477 nm, what is the concentration of [FeSCN]^{2+} in the solution? (Assume the path length is 1 cm)

1 answer:

alexandr1967 [171]3 years ago

8 0

Answer:9.81×10^-5 M

Explanation: See attachments

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concentrated phosphoric acid is90% H3PO4 by mass and the remaining mass is water. The molarity of H3PO4 is 12.2M at temperature

blondinia [14]

Answer:

82.0 mL

Explanation:

Step 1: Given data

Concentration of concentrated acid (C₁): 12.2 M

Volume of concentrated acid (V₁): ?

Concentration of dilute acid (C₂): 1.00 M

Volume of dilute acid (V₂): 1.00 L

Step 2: Calculate the required volume of the concentrated acid

We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL

4 0

2 years ago

A 25 kg rock is placed in a graduated cylinder with water.the volume of the fluid is 18.3ml.calculate the density of the rock in

ipn [44]

Answer:

=> 1366.120 g/mL.

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m) = 25 Kg

Volume (v) = 18.3 mL.

From our question, we are to determine the density (rho) of the rock.

The formula:

$p = \frac{m}{v}$

First let's convert 25 Kg to g;

1 Kg = 1000 g

25 Kg = ?

$= \frac{25 × 1000}{1}$

= 25000 g

Substitute the values into the formula:

$p = \frac{25000 g}{18.3 ml}$

= 1366.120 g/mL.

Therefore, the density (rho) of the rock is 1366.120 g/mL.

8 0

2 years ago

At which temperature and pressure will a sample of neon gas behave most like an ideal gas?

Andreas93 [3]

Answer:

At STP, 760mmHg or 1 atm and OK or 273 degrees celcius

Explanation:

The standard temperature and pressure is the temperature and pressure at which we have the molecules of a gas behaving as an ideal gas. At this temperature and pressure, it is expected that the gas exhibits some properties that make it behave like an ideal gas.

This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable

Take for instance, one mole of a gas at stp occupies a volume of 22.4L. This particular volume is not obtainable at other temperatures and pressures but at this particular temperature and pressure. One mole of a gas will occupy this said volume no matter its molar mass and constituent elements. This is because at this temperature and pressure, the gas is expected to behave like an ideal gas and thus exhibit the characteristics which are expected of an ideal gas

4 0

3 years ago

Read 2 more answers

In a 0.737 m solution, a weak acid is 12.5% dissociated. calculate ka of the acid.

r-ruslan [8.4K]

Hello! Let me try to answer this :)

Thanks and please correct if there are any mistakes ^ ^

8 0

3 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

<u>Answer:</u> The theoretical yield of acetanilide is 6.5 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For aniline:</u>

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

<u>For acetic anhydride:</u>

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

3 years ago