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2 years ago

The stalk that holds the anther up so that pollination and fertilization can occur is the _______. :

Chemistry

2 answers:

Andrews [41]2 years ago

6 0

The filament holds up the anther so that pollination and fertilization can occur!

kap26 [50]2 years ago

3 0

The answer is filament. Hope that helped!

You might be interested in

What is the metal cation in K,SO,? (cation is pronounced cat-ion and refers to a positively charged ion)

slamgirl [31]

Answer:

Potassium cation = K⁺²

Explanation:

The metal cation in K₂SO₄ is K⁺². While the anion is SO₄²⁻.

All the metals have tendency to lose the electrons and form cation. In given compound the metal is potassium so it should form the cation. The overall compound is neutral.

The charge on sulfate is -2. While the oxidation state of potassium is +1. So in order to make compound overall neutral there should be two potassium cation so that potassium becomes +2 and cancel the -2 charge on sulfate and make the charge on compound zero.

2K⁺² , SO₄²⁻

K₂SO₄

8 0

2 years ago

Which of these have the same number of particles as 1 mole of water H2O

Fed [463]

Answer:

It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.

Explanation:

The question is not very much clear.

If you are asking for molecules then 1 mole water= 6.023 * 10^23

If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3

If you are asking for particles then,

So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.

I hope that was helpful!

H=1 proton,1 electron

O=8 protons,8 neutrons and 8 electrons

total particles in one H2O molecule-28

total no. of particles in 1 mole of water- 6.023 * 10^23 * 28

8 0

2 years ago

A rock is an

Gekata [30.6K]

Answer:

mineral

Explanation:

6 0

3 years ago

Read 2 more answers

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago

Write out a balanced, molecular equation, total ionic equation, and net ionic equation for each:

dangina [55]

Answer:

1:MgCO3 (s) + 2 NaNO3

2:agcl(s)+kno3(aq)

3:alcl3(aq)+h2(g)

4:NaNO3 + CO2 + H2O

(not sure abt last one)

6 0

3 years ago