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3 years ago

Which of the following examples could be considered a genetically modified food source?

1 answer:

4 0

“Carrot seeds are injected with a substance derived from parsnips that makes them more resistant to rot” is considered as genetically modified food source.

Option: C

Explanation:

"Genetically modified foods" (GM-Foods) also called "bioengineered food" which is considered different from traditional cross breeding. These foods are developed when an organism carry changes which are injected into their DNA by using genetic engineering method. USA is more developing and fast to launch GM-Foods for example researchers at Texas A&M AgriLife's vegetable and Fruit improvement center observed that calcium absorption is more when consumption is done through carrot. They observed 41% more calcium absorption than regular carrot intake.

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For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+

Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0

3 years ago

Particle quantity/number of Cu(NO3)2

Murrr4er [49]

Answer:

Cupid nitrate is what I'm going for

8 0

3 years ago

Is this correct im so confused.

inysia [295]

Answer:

the last option!!!!

Explanation:

because its right

4 0

3 years ago

The second-order diffraction for a gold crystal is at an angle of 22.20o for X-rays of 154 pm. What is the spacing between the c

Alenkinab [10]

Answer: The spacing between the crystal planes is $4.07\times 10^{-10}m$

Explanation:

To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:

$n\lambda =2d\sin \theta$

where,

n = order of diffraction = 2

$\lambda$ = wavelength of the light = $154pm=1.54\times 10^{-10}m$ (Conversion factor: $1m=10^{12}pm$ )

d = spacing between the crystal planes = ?

$\theta$ = angle of diffraction = 22.20°

Putting values in above equation, we get:

$2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m$

Hence, the spacing between the crystal planes is $4.07\times 10^{-10}m$

4 0

3 years ago

How many neutrons are in 7.00 g of 13c?

Minchanka [31]

You multiply the number of atoms by 12 to get how many electrons (since each atom has 12 electrons in it)
you multiply the number of atoms by 13 to get how many neutrons
(since each atom of this isotope has 13 neutrons in it)

5 0

3 years ago