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Rama09 [41]
3 years ago
6

if the chromosome number in the sporophyte generation of a plant is 32 what is its number in the gametophyte generation?

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
3 0

Chromosomes have hereditary function that are capable of self-duplication and they have thread-like structures seen inside the nucleus of plant and animal. In plants, they go through a cycle called alteration of generation that has multicellular stages. One is sporophyte that is a diploid, and the other is gametophyte with a haploid.

<span>Moreover, spores are produce through meiosis so it go through a process of chromosome reduction in order to create a haploid spores. While, gametes undergo mitosis and the structure is already a haploid, which means that the number of chromosomes will not change in order to produce haploid. In this case, the plant have both spores and gametes and the similarities is that, they are both singles celled and a haploid. Therefore, the number of the chromosome in the gametophyte generation is also 32.</span>

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Using wolfram alpha or some other reference, determine which of these elements would be liquid at 525 k (assume samples are prot
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3 years ago
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Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

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Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

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Thus;

n = Initial volume/22.4L

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