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galben [10]
3 years ago
12

What are the kingdoms of eukaryotes

Chemistry
1 answer:
Talja [164]3 years ago
4 0
Protists,fungi,plantar and animalia
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Calculate the mass of 3 moles of clorine​
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Signals from the make skeletal muscles move. A. Nervous system B.
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Signals from the A. Nervous system make the skeletal muscles move.

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3 years ago
Three other application of Bernoulli's principle ?
Makovka662 [10]
<h2><u>Answer:</u></h2>

Bernoulli's Theorem in a general sense relates the weight, speed, and rise in a moving fluid (liquid or gas), the compressibility and consistency (internal grinding) of which are insignificant and the flood of which is predictable, or laminar.

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(2): Vena Contracta: The fluid stream from gap contracts at a separation minimal outside the opening to a neck, called Vena Contracta.

The territory of cross-segment of a fly is littler than a zone of opening. From this reality, we can discover the coefficient of withdrawal.

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7 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
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