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andrey2020 [161]
3 years ago
5

Which tissue provides medical strength to plant​

Chemistry
1 answer:
blondinia [14]3 years ago
3 0

Answer:

mechanical tissue

Explanation:

hahhshshs

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Calculate the mass of 1 molecule of water and 100 molecules of glucose​
lesya692 [45]

Explanation:

Mass of one atom of water = Mass of 1 mole / Avogadro's number

Mass of one atom of water = 18g/mol / 6.022×10²³molecules

Mass of one atom of water = 2.989 × 10^-23 g

Mass of glucose = 180(one molecule)

Mass of 100 molecules of glucose = 180×100 =18000

3 0
3 years ago
based on distance to the nearest water well, which drilling site would pose the least risk of methane contamination?
Marysya12 [62]

Based on distance to the nearest water well, the drilling site that would pose the least risk of methane contamination would be the one farthest away from the water well.

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The atmosphere is the air that surrounds us. It is a mixture of gases, including nitrogen, oxygen, and carbon dioxide. The atmosphere protects us from the Sun's ultraviolet radiation and from the cold of space. It also contains the Earth's weather, which is caused by the Sun's energy.

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7 0
2 years ago
** will mark you brainliest if answered correctly **
lord [1]

Answer:

D. 23.9 liters

Explanation:

  • Molarity is defined as the no. of moles of solute dissolved in 1.0 L of the solution.

M = (no. of moles)solute / (V of the solution (L))

M = (mass/molar mass)NaCl / (V of the solution (L))

M = 0.03 M, mass of NaCl = 42.0 g, molar mass = 58.44 g/mol.

∴ V of the solution = (mass/molar mass)NaCl / M = [(42.0 g/58.44 g/mol)] / (0.03 M) = 23.96 L.

<em>So, the right choice is: D. 23.9 liters</em>

7 0
3 years ago
The dipole measured for HI is 0.380 D. The bond distance is 161 pm. What is the percent ionic character of the HI bond?
Feliz [49]

Answer:

Percent ionic character of HI bond is 4.91%.

Explanation:

<h3>Given Data:</h3>

Measured Dipole = 0.380D

bond distance = d = 161pm = 1.61*10^-8 cm

<h3>Calculation:</h3>

% ionic character is determined by following equation:

% ionic= (dipole measured/dipole calculated)*100

Now,

Dipole(calc)=qd

Dipole(calc)= (1.6*10^{-19}*3*10^{9})esu  *1.61*10^{-8}cm

(In above step 3*10^8 is multiplied to convert coulomb into esu)

Dipole(calc)=7.728*10^{-18} esu*cm

As,

10^{-18}esu*cm= 1D

So,

Dipole(calc)=7.728D

Now we can % ionic character using above equation:

%ionic=(0.380D/7.728D)*100

% ionic character=4.91%

5 0
4 years ago
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