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3 years ago

All of the following are examples of human organs EXCEPT the

Chemistry

2 answers:

mestny [16]3 years ago

8 0

Answer:

eyelashes are not a part of the the human organs

Allushta [10]3 years ago

4 0

All of the following are examples of human organs EXCEPT the eyelashes.

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G. Name a separating method for each of the question. (3)

erik [133]

Answer:
1) Evaporation
2) Distillation
3) Mix them in water as salt is water soluble but sand isn’t and then filter

Hope this helps! :)

6 0

2 years ago

Which quantity is a vector quantity?

yan [13]

The answer is accleration

6 0

2 years ago

A buffer solution contains 0.368 M hydrocyanic acid and 0.360 M potassium cyanide . If 0.0513 moles of sodium hydroxide are adde

MrRa [10]

Solution :

Millimoles of hydrocyanic acid = 225 x 0.368

= 82.8

Millimoles of potassium cyanide = 225 x 0.360

= 81

Millimoles of sodium hydroxide = 51.3

Therefore,

pOH = pKb + log [salt - C / bas + C]

= 4.74 + log[82.8 - 51.3 / 81 + 51.3]

= 4.102

Therefore, pH = 9.05

7 0

3 years ago

In an experiment, a student dissolves 56.3g of MgCl2 in 0.12kg of solvent. What is the molality of this solution? Show your work

Kazeer [188]

<u>Answer:</u> The molality of magnesium chloride is 4.93 m

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}}{M_{solute}\times W_{solvent}\text{ in kg}}$

Where,

$m_{solute}$ = Given mass of solute $(MgCl_2)$ = 56.3 g

$M_{solute}$ = Molar mass of solute $(MgCl_2)$ = 95.2 g/mol

$W_{solvent}$ = Mass of solvent = 0.12 kg

Putting values in above equation, we get:

$\text{Molality of }MgCl_2=\frac{56.3}{95.2\times 0.12}\\\\\text{Molality of }MgCl_2=4.93m$

Hence, the molality of magnesium chloride is 4.93 m

7 0

3 years ago

Suppose 13.6 g of barium nitrate is dissolved in 300. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final mol

Elis [28]

Answer:

The molarity of barium cation in the solution is 0.173 M

Explanation:

Step 1: The balanced equation

Ba(NO3)2(aq) + Na2CrO4 (aq) → BaCrO4(s) + 2NaNO3(aq)

Step 2: Data given

Mass of Barium nitrate = 13.6 grams

Volume of 0.40M sodium chromate = 300 mL

Step 3: Calculate moles of Ba(NO3)2

Moles = mass / molar mass

Moles = 13.6 grams / 261.34 g/mol

Moles = 0.052 moles

Step 4: Calculate moles of Na2CrO4

Moles = Molarity * Volume

Moles Na2CrO4 = 0.40 * 0.3L

Moles Na2CrO4 = 0.12 moles

Step 5: Calculate limiting reactant

Na2CrO4 is in excess so all of Ba(NO3)2 will be consumed and reacts to form BaCrO4(s) in the form Ba2+

Step 6: Calculate moles of Ba2+

n(Ba2+)=n(BaCrO4) =n(Ba(NO3)2 = 0.0520 moles

Step 7: Calculate molarity of Ba2+

C=n/v so C(Ba2+)=0.0520/0.300 = 0.173 M

The molarity of barium cation in the solution is 0.173 M

4 0

3 years ago