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Paha777 [63]
3 years ago
7

A mutation occurs in a gene causing a pea plant produce 75% fewer leaves. this would be considered what kind of mutation?

Chemistry
1 answer:
denpristay [2]3 years ago
5 0

Answer:

negative

Explanation:

There is decrease of a character in the new progeny so it is negative mutation

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Convert the following to Celsius
Naya [18.7K]

9. (80-32)5/9

48×5/9

240/9

26.66°C

9. (90-32)5/9

58×5/9

32.22°C

10 (212-32)5/9

180×5/9

20×5

100°C

8 0
3 years ago
Which cell structure is correctly paired with its primary function?
Butoxors [25]

Answer:

C

Explanation:

Mitochondria provides energy (ATP) for cells to use

Nucleus provides genetic code (DNA)

Ribosomes assemble amino acids chains based on the DNA from nucleus to form proteins

6 0
3 years ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2
Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

5 0
2 years ago
The density of aluminum is 2.70 g/mL. If you use 108 g of aluminum foil, what is its volume in mL?
Natali5045456 [20]

Answer:

volume =40.0 ml

Explanation:

density = mass/ volume

volume = mass/ density

volume = 108/ 2.70

volume =40.0 ml

3 0
2 years ago
A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency
saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

Frequency of photon = ?

Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J =  6.63×10⁻³⁴ Js ×f

f =  5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

4 0
2 years ago
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