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nexus9112 [7]
2 years ago
7

The gene for tallness (T) in a pea plant is dominant over the gene for shortness (t).

Chemistry
2 answers:
astra-53 [7]2 years ago
6 0

Answer:

C. 75%

Explanation:

From a heterozygous cross (Tt x Tt) the produced offspring would consist of:

  • 25% TT -it would be tall-.
  • 50% Tt -it would be tall as the gene T is dominant-.
  • 25% tt -it would not be tall-.

Thus the produced offspring that would be tall is (50% + 25%) 75%. The answer is option C.

Grace [21]2 years ago
4 0

Answer:

C

Explanation:

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50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
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8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

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$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

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                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

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$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

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pOH =- $\log[OH^-]$

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Now, since pH + pOH = 14

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