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SOVA2 [1]
3 years ago
5

Take a beaker (jar or jug) with some water and put some drops of red or blue ink in it.Take a tender twig of any flowery plant w

ith leaves and flowers ,preferably white flowers,and put it in the beaker for 6 to 8 hours. What do you observe?For further examination cut across its stem and look for the presence of coloured water.Explain your observations.
Chemistry
1 answer:
frutty [35]3 years ago
8 0

Answer:

After about 6 to 8 hours, the white flowers begins to develop red spots near the edge of the petals.

When the stems were cut, coloured water was observed along the length of the stem.

Explanation:

In the experiment, red ink was placed in the beaker of water and a red-coloured water was obtained. Then the tender twig of a flowery plant (e.g. carnation) with leaves and white flowers was placed in the beaker. After about 6 to 8 hours, the white flowers began to develop red spots near the edge of the petals. When the stems were cut, coloured water was observed along the length of the stem. The observations in this experiment is due to the fact that the stem conducted the water and the ink through the xylem up to the leaves. On getting to the leaves, the water evaporates from the leaves in a process known as transpiration leaving behind the red ink. This ink then appears as the red spots on the leaves.

The process of transpiration in the leaves results in the continuous conduction of water from the stem to leaves, and hence, the appearance of more coloured spots with time.

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when Mn2 ions are separated from the mixture, they go through a series of oxidizing and reducing steps. Write the reaction equat
goldfiish [28.3K]

Answer: hello some part of your question is missing below is the missing part

when H₂O and H₂O₂ is added to Mn(OH)₂(s) and put in water bath to dissolve

answer : attached below

Explanation:

When Mn²⁺ ions are separated from the mixture, attached below are the requires reaction equations that shows the process of separation.

Mn²⁺ ions are separated to the right of the reaction  equations

8 0
3 years ago
If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?
Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L  - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

new volume -  125 mL + 25 mL = 150 mL

therefore new molarity is

c = 0.01875 mol / 0.150 L  = 0.125 M

answer is 0.125 M

7 0
3 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
How many helium atoms are there in a helium blimp containing 539kg of helium
MaRussiya [10]
Atomic mass of helium is 4.002642g/mol 
(542000g)/(4.002642g/mol)*6.02*10^23 = 8.15*10^28 atoms
3 0
3 years ago
Read 2 more answers
1. By means of orbital diagrams, write down the
Dvinal [7]

Answer:

a) 1s22s22p63s1

b) 3s² 3p¹

c) 1s22s22p63s23p3

d) 3s² 3p⁵

e) 3s2

f) 1s22s22p63s23p2

g) 1s22s22p63s23p4

h) 3s² 3p⁶

3 0
3 years ago
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