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mezya [45]
3 years ago
7

A bumblebee helps plants reproduce by carrying pollen from a flower on one plant to a flower on another plant. Which reproductiv

e process is this most similar to?
Chemistry
1 answer:
Oliga [24]3 years ago
8 0

Answer:

cross pollination

Explanation:

this is the transfer of pollen grains from the anther of a flower to the stigma of another flower of the same kind.

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D.

atomic modifier

Explanation:

. Identify at least three reasons the Articles of Confederations failed as a governing document. In your opinion, evaluate which defect was most debilitating, using evidence and your knowledge of American government to justify your position. [5]

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3 years ago
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Explain why the dissolved component does not settle out of a solution
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3 years ago
Why is this flower able to self-pollinate
Andrei [34K]

Answer:

A flower is self-pollinated if pollen is transferred to it from any flower of the same plant and cross-pollinated if the pollen comes from a flower on a different plant.

Explanation:

This is because each flower is capable of fertilizing itself by autogamy. Autogamy means that the male part of a flower sends pollen to the female part of the same flower.

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3 years ago
When uranium decays inside Earth, what does it produce?
noname [10]

Answer:

New rocks

Explanation:

8 0
2 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
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