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Nookie1986 [14]
3 years ago
6

Indicate whether the following statements about genetically modified foods are true or false. Genetically modified foods are res

istant to insects and plant viruses. Genetically modified foods are not yet available to consumers at the grocery store. Genetically modified foods have delayed fruit ripening for longer shelf life during transportation. Genetically modified foods have enhanced flavor and nutritional content.
Biology
2 answers:
GREYUIT [131]3 years ago
6 0
Genetically  modified foods are  resistant  to  insents  and  plant  viruses  -  TRUE

Genetically   modified  foods  are  yet  available to  consumers at  the  grocery  store    -false

genetically  modified  foods  have  delayed  fruit  ripening  for  shell  life  during  transportation  -TRUE

Genetically  modified  foods  have  enhanced   flavor  and  nutritional  content -True
Serjik [45]3 years ago
5 0

Answer:

True,False,True,True

just answered it

Explanation:

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Be sure to answer all parts.
Aleks04 [339]

Answer:

pH = 3.215

Explanation:

From the given information;

Using the equation for the dilution of  a stock solution:

Since moles of C5H5N = moles of HNO3

Then:

M_{C_5H_5N}\times V_{C_5H_5N}= M_{HNO_3}\times V_{HNO_3}

0.0750 M \times 2.65 L = 0.407 M \times V_{HNO_3}

V_{HNO_3}= \dfrac{0.0750 M \times 2.65 L}{0.407 M}

V_{HNO_3}=488.33 \ mL

The reaction between C5H5NH and H2O is as follows:

C_5H_5N^+H + H_2O ⇄ C_5H_5N + H_3O^+

Molarity  \ of \ C_5H_5N^+H = \dfrac{moles}{addition \ of\  the\  total\ volume}

Molarity  \ of \ C_5H_5N^+H = \dfrac{0.0750 \ M \times 2.65 \ L}{2.65 \ L + 0.48833 \ L}

Molarity  \ of \ C_5H_5N^+H = \dfrac{0.19875 \ ML}{3.13833 L }

= \ 0.06333\ M

Now, the next step is to draw out the I.C.E table.

    C_5H_5N^+H + H_2O ⇄ C_5H_5N + H_3O^+

I       0.06333                      0          0

C        - x                               x          x  

E      0.06333 -x                   x          x

K_a  = \dfrac{10^{-14}}{1.7 \times 10^{-19}} \\ \\  K_a = 5.8824 \times 10^{-6}

K_a = \dfrac{[C_5H_5N][H_3O^+] }{[C_5H_5N^+H] }  \\ \\  5.8824 \times 10^{-6} = \dfrac{x^2}{0.06333 - x}

Assuming x < 0.06333

x^2 = 5.8824 \times 10^{-6} \times 0.06333

Then

x^2 = 3.72532392 \times 10^{-7} \\ \\ x= \sqrt{3.72532392 \times 10^{-7}} \\ \\  x = 6.1035 \times 10^{-4} \ M

[H_3O^+] =  x = 6.1035 \times 10^{-4} \ M  \\ \\  pH = -log (6.1035 \times 10^{-4}) \\ \\ \mathbf{\\ pH = 3.215}

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