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2 years ago

Example diluted solution? 1.2.3 .4.5.

Chemistry

1 answer:

zhuklara [117]2 years ago

6 0

Answer:

dilution

Explanation:

a solution which contains small amount of solute dissolve in large amount of solvent is called dilutEd solution.. dilutEd solution is prepared when a small amount of solute is dissolve in large amount of solvent..

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Which of the four macromolecules are the largest

Kruka [31]

Lipid you can look up the chemical structures and lipids are the longest which make them the largest

5 0

2 years ago

A chemist dissolves 867. mg of pure barium hydroxide in enough water to make up 170. mL of solution. Calculate the pH of the sol

Ilya [14]

Answer: The pH of the solution is 11.2

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Ba(OH)_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.867g}{171g/mol}=0.00507mol$ (1g=1000mg)

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.00507\times 1000}{170}$

$Molarity=0.0298$

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pOH=-\log [OH^-]$

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}$

According to stoichiometry,

1 mole of $Ba(OH)_2$ gives 2 mole of $OH^-$

Thus 0.0298 moles of $Ba(OH)_2$ gives = $\frac{2}{1}\times 0.0298=0.0596$ moles of $OH^-$

Putting in the values:

$pOH=-\log[0.0596]=2.82$

$pH+pOH=14$

$pH=14-2.82$

$pH=11.2$

Thus the pH of the solution is 11.2

8 0

2 years ago

At 0 degrees Celsius, a gas occupies 22.4L. How hot must the gas be in celcius to reach a volume of 25.0L

NikAS [45]

Answer:

31.7 °C

Explanation:

Charles law states that for volume of a gas is directly proportional to the absolute temperature for a fixed amount of gas at constant pressure

we can use the following equation

V1/T1 = V2/T2

where V1 is volume and T1 is temperature at first instance

V2 is volume and T2 is temperature at second instance

temperature should be in kelvin scale

T1 - 0 °C + 273 = 273 K

substituting the values in the equation

22.4 L / 273 K = 25.0 L / T2

T2 = 304.7 K

temperature in celcius is - 304.7 K - 273 = 31.7 °C

the gas must be 31.7 °C to reach a volume of 25.0 L

7 0

2 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

A compound has a pka of 7.4. to 100 ml of a 1.0 m solution of this compound at ph 8.0 is added 30 ml of 1.0 m hydrochloric acid.

Monica [59]

Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.

6 0

3 years ago

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Example diluted solution? 1.2.3 .4.5.​

Example diluted solution? 1.2.3 .4.5.