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choli [55]
2 years ago
15

Which of the following is an example of a genetically modified crop?

Chemistry
2 answers:
Aloiza [94]2 years ago
5 0

Wheat that grows 20 percent more wheat

lord [1]2 years ago
4 0

Answer:

Explanation:

Genetically modified crops (GM crops) are plants used in agriculture, the DNA of which has been modified using genetic engineering methods. In most cases, the aim is to introduce a new trait to the plant which does not occur naturally in the species.

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Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce
levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca^{2+}(aq) and 0.0390 M Ag^{+}(aq). What will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate? What percentage of the Ca^{2+}(aq) can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

      Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}

[Ag^{+}] = 0.0390 M

When Ag_{2}SO_{4} precipitates then expression for K_{sp} will be as follows.

         K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]

        1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]

       [SO^{2-}_{4}] = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

         CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}

      K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

     4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788

           [Ca^{2+}] = 0.00625 M

Now, we will calculate the percentage of Ca^{2+} remaining in the solution as follows.

               \frac{0.00625}{0.05} \times 100

                 = 12.5%

And, the percentage of Ca^{2+} that can be separated is as follows.

                     100 - 12.5

                     = 87.5%

Thus, we can conclude that 87.5% will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate.

4 0
2 years ago
Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina
yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0
3 years ago
A galaxy contains stars, gas, and dust. This matter must be held together or the galaxy will break apart. What is responsible fo
Marrrta [24]
Gravity is responsible !
5 0
2 years ago
Consider the temperature versus time graph below. I which region is the substance in both the solid phase and the liquid phase?
SVEN [57.7K]

Answer:

2

Explanation:

The change of state occurs at a constant temperture and pressure. In the grahp we can see while the time passes, the temperature doesn't change.

The rect number 4 correspond to a liquid-gas phase

8 0
3 years ago
Read 2 more answers
The SI unit of pressure is the ______________.
steposvetlana [31]

The answer is Pascal

5 0
3 years ago
Read 2 more answers
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