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katen-ka-za [31]
1 year ago
8

A biochemical engineer isolates a bacterial gene fragment and dissolves a 10.0-mg sample in enough water to make 30.0 mL of solu

tion. The osmotic pressure of the solution is 0.340 torr at 25°C. (a) What is the molar mass of the gene fragment?
Chemistry
1 answer:
Novay_Z [31]1 year ago
7 0

The molar mass of the gene fragment is 19182 g/mol.

What is osmotic pressure ?

Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in a pure solvent by osmosis. Potential osmotic pressure is the maximum osmotic pressure that could develop in a solution if it were separated from its pure solvent by a semipermeable membrane.

We employ the osmotic pressure equation to determine the solute's concentration, which is:

π = iMRT

Using the values in the equation above, we obtain:  19182 g/mol.

To learn more about gene fragment click on the link below:

brainly.com/question/22426204

#SPJ4

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A scientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. How many grams of na3po4
natima [27]

Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given  : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 \frac{mol}{L}

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL  

Volume of solution = 725 mL * \frac{1L }{1000 mL}

Volume of solution = 0.725 L


Following steps can be  done to find mass of  :

<u>Step 1 : Write the dissociation reaction of Na₃PO₄ .</u>

Na_3PO_4    3 Na^+  +  PO_4^3^-

<u>Step 2:  Find moles of Na⁺ ions : </u>

Mole of Na⁺ ions can be calculated using molarity formula  which is :Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }

Plugging value of Molarity and volume

1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}

Multiplying both side by 0.725 L

1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L

<em>Mole of Na⁺ ions = 0.725 mol</em>

<u>Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :</u>

Mole ratio  is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion =  3

<em>Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3 </em>

<u>Step 4 : To find mole of Na₃PO₄ </u>

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

Mole of Na_3PO_4= Mole of Na^+  * Mole ratio

Mole of Na_3PO_4 = 0.725 mol  *  \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }

<em>Mole of Na₃PO₄ =  0.242 mol </em>

<u>Step 5 : To find mass of Na₃PO₄</u>

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using  molar mass of Na₃PO₄ as :

Mass (g) = mole (mol) * molar mass \frac{g}{mol}

Mass of Na_3PO_4 =  0.242 mol * 163.94 \frac{g}{mol}

Mass of Na₃PO₄ = 39.619 g

<u>Step 6 : To round off mass of Na₃PO₄ to correct sig fig .</u>

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g


8 0
2 years ago
1.1 Outline a method for separating the chalk from potassium chloride,
miss Akunina [59]

Answer:

Explanation:

To separate the a mixture of chalk and potassium chloride, we must not that chalk is calcium carbonate compound, CaCO₃.

Calcium carbonate has low solubility in water. KCl is readily soluble in water and it is also an ionic compound.

To separate a mixture of compounds with various solubility, we can carryout dissolution, filtration and evaporation.

We first pour pure water into the mixture. Water will dissolve the potassium chloride readily.

Then using a filter paper we filter out the suspended chalk particles. Leave the filtrate to then dry and collect it.

The solution filtered should be evaporated to dryness. This will leave the KCl behind from the solution.

5 0
3 years ago
The atomic number is the number of protons in the atom of the element
ipn [44]

Answer:

True

Explanation:

If you look closley at the nucleus, you don't count the neutrons just the prtons which then effect the electrons.

Good luck :)

5 0
3 years ago
Teeth, tongue, throat, stomach, small intestines, large intestines, and rectum are all part of ___________ system during the sys
const2013 [10]
The answer is B. Gastrointestinal.
3 0
3 years ago
Read 2 more answers
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
2 years ago
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