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2 years ago

What compounds can you find in a whole wheat bread

2 answers:

6 0

Apigenin-6-C-arabinoside-8-C-galactoside. ...
Apigenin-6-C-galactoside-8-C-arabinoside. ...
9,12,13-Trihydroxy-trans-10-octadecenoic acid (pinellic acid)

givi [52]2 years ago

5 0

Explanation:

Three major bitter compounds in whole wheat bread crumb were identified.

...

Apigenin-6-C-arabinoside-8-C-galactoside. ...

Apigenin-6-C-galactoside-8-C-arabinoside. ...

9,12,13-Trihydroxy-trans-10-octadecenoic acid (pinellic acid)

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he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o

frosja888 [35]

<u>Answer:</u> The number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

<u>Explanation:</u>

We are given:

$K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L$

To calculate the concentration of a substance, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ ......(1)

Concentration of ammonia:

$[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L$

Concentration of ammonium iodide:

$[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L$

For the given chemical reaction:

$NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI][NH_3]}{[NH_4I]}$

Putting values in above equation, we get:

$7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}$

$[HI]=2.53\times 10^{-4}$

Calculating the moles of hydrogen iodide by using equation 1, we get:

$2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}$

Hence, the number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

3 0

2 years ago

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is no

nasty-shy [4]

This question is incomplete, the complete question is;

Tonksite is a solid at 300.00K. At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.

Answer: the sublimation pressure of the solid at the melting point is 0.3727 atm

Explanation:

Given that;

T1 = 300 K

T2 = 400 K

H_sub = 66 kJ/mol = 66000 J/mol

P1 = 5.00 × 10⁻⁴ atm

p2 = ?

now using the expression

log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R × 2.303 ) (( T2 - T1) / T1T2)

now we substitute of given values into the expression

log(p2/p1) = (66000 / 8.314 × 2.303 ) (( 400 - 300) / 300 × 400 )

p2 = 0.3727 atm

therefore the sublimation pressure of the solid at the melting point is 0.3727 atm

6 0

3 years ago

The process in which organisms grow and replace worn-out cells is called: A. Cell regeneration B. Cell division C. Mitosis D. Bo

djyliett [7]

The mitosis is the cell division so it both B and C

7 0

3 years ago

Which description best characterization the motion of particles in a solid

marta [7]

Are there any choices? Because from what the question is it seems like we need choices to help

4 0

3 years ago

Read 2 more answers

+ H₂O

trapecia [35]

Answer:

None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.

Explanation:

Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.

Before balanced Left side.

Cl-2

O-8

H-2

Before balanced right side.

H-1

Cl-1

O-3

That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.

(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)

8 0

2 years ago