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2 years ago

4 main parts of a flower Include the function of each pls help me

Chemistry

1 answer:

ZanzabumX [31]2 years ago

6 0

Answer:

Calyx or sepals
Corolla
Androecium
Gynoecium

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The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2

NNADVOKAT [17]

PbCl₂(aq) → Pb²⁺(aq) + 2Cl⁻(aq)

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Pb²⁺(aq) + 2Cl⁻(aq) ⇄ PbCl₂(s)

At increase the concentration of chloride ions - concentration of lead ions decreases, the lead chloride is formed.

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3 years ago

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What is the correct equilibrium constant

Ksju [112]

Answer:

(CO2)+(CF)/ (COF)².

Explanation:

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3 years ago

What is the main cause of an earthquake

Natali5045456 [20]

Answer:

I thinks it's stress caused by plate movement hope this helps

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3 years ago

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For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

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3 years ago

The work function is the energy that must be supplied to cause the release of an electron from a photoelectric material. The cor

vladimir2022 [97]

Answer:

No photoelectric effect is observed for Mercury.

Explanation:

From E= hf

h= Plank's constant

f= frequency of incident light

Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23

f= 11×10^14 Hz

The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface

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3 years ago