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4 years ago

Most types of commercial flour are sifted

Chemistry

1 answer:

zubka84 [21]4 years ago

3 0

Answer:

It improves the texture of the flour

Explanation:

Flour is supposed to have a very fine texture. Commercial flour is not expected to occur in lumps. Lumps reduce the utility of the flour.

Hence, to improve the texture of the commercial flour, it is thoroughly sifted to ensure that there are no large lumps in the flour.This is so important when producing cake flour where the achievement of fine texture is of utmost priority.

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What is the charge of CO3?

Tema [17]

CO3 has an overall charge of -2. Within CO3, any carbon atom bonds with a double bond to oxygen atoms as well as single bonds to double oxygen atoms.

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3 years ago

DOES YOUR EARS HAVE BONES?

inna [77]

Yes, they do!
Ear bone, also called Auditory Ossicle, any of the three tiny bones in the middle ear of all mammals. These are the malleus, or hammer, the incus, or anvil, and the stapes, or stirrup

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3 years ago

Calculate the percent dissociation of trimethylacetic acid(C6H5CO2H) in a aqueous solution of the stuff.

Sophie [7]

Answer:

1.112%.

Explanation:

Step one : write out the correct acid dissociation reaction. This is done below:

C6H5CO2H <=====> C6H5CO2^- + H^+.

At equilibrium, the concentration of C6H5CO2H = 0.5 M - x and the concentration of C6H5CO2^- = x and the concentration of H^+ = x.

The ka for C6H5CO2H = 6.3 × 10^-5.

Step two: find the value for the concentration of C6H5CO2^- and H^+. This can be done by using the equilibrium dissociation Constant formula below;

Ka = [C6H5CO2^- ] [H^+] / C6H5CO2H.

ka = [x] [x] / [0.5 - x].

6.3 × 10^-5 = (x)^2 / [0.5 - x].

x^2 = 3.15 ×10^-5 - 6.3 × 10^-5 x.

x^-2 + 6.3 × 10^-5 x - 3.15 × 10^-5.

Solving for x we have x= 0.0055632289702004 = 0.00556.

Therefore, at equilibrium the concentration of H^+ = 0.00556 M and the concentration of C6H5CO2H= O.5 - 0.00556 = 0.494 M.

Step three: Calculate the equilibrium pH. This stage can be ignored for this question since we are not asked to calculate for the pH.

The formular for pH = - log [H^+].

pH= - log [0.00556].

pH= 2.255.

Step four: find the percentage dissociation.

Percentage dissociation = ( [H^+] at equilibrium/ [ C6H5CO2H] at Initial concentration ) × 100%.

Percentage dissociation=( 0.00556/ 0.5 ) × 100.

=Percentage dissociation= 1.112%.

3 0

4 years ago

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$