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dimulka [17.4K]

3 years ago

5

The mutchkin and the noggin a mutchkin is a scottish unit of liquid measure equal to 0.42 l. How many mutchkins are required to

fill a container that measures one foot on a side?

2 answers:

I am Lyosha [343]3 years ago

8 0

Answer:

$67.41Mutchkin$

Explanation:

Hello,

In this case, the equivalence is:

$1Mutchkin=0.42L$

Thus, the volume of the container is based on the given one foot on a side as:

$V=1ft^3$

In addition, the equivalence between cubic feet and liters is:

$1ft^3=28.317L$

Thus, the mutchkins turns out:

$1ft^3*\frac{28.317L}{1ft^3} *\frac{1Mutchkin}{0.42L} =67.41Mutchkin$

Best regards.

Rasek [7]3 years ago

6 0

Given that 1 mutchkin (a scottish unit) = 0.42 L

Now the volume of container (assuming it to be cubical, as only one side is given) = a^3 = 1^3 = 1ft^3

Now 1 foot = 3.048 dm [1 decimeter= 0.1 m]

Hence 1 ft X 1ft X 1ft = 3.048 X 3.048 X 3.048 dm^3 = 28.32 dm^3

We know that 1L = 1 dm^3

therefore volume of container in Litres = 28.32 L

now as given that 0.42 L = 1 mutchkin

therefore 1 L = 1 / 0.42 mutchkin

28.32 L = 28.32 / 0.42 mutchkin = 67.43 mutchkins [This much mutchkin is required to fill the container]

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3 years ago

Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,

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Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

CH₃CH₂CO₂H = HA

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pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

(a)

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6 0

4 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

<u>Part (a) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

<u>Part (b) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago

Cytokinesis happens differently for plant and animal cells. Both separate cytoplasm between two new daughter cells. However, whi

victus00 [196]

Answer:

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Explanation:

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The process of cytokinesis is different for animal cell and plant cell.

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Whereas, in case of animal cell, the plasma membrane contracts itself along the center, until the two daughter cells are separated from each other.

Hence, from the given information of the question,

The correct answer is animal cell.

7 0

3 years ago