The osmotic pressure, π, of a solution of glucose is 132 atm . find the molarity of the solution at 298 k.

Answers

Answer 1

Answer:

The formula for osmotic pressure is:

$\Pi = iMRT$

where $\Pi$ is osmotic pressure, $i$ is van't Hoff's factor, $M$ molarity, $R$ is Ideal gas constant, and T is Temperature.

$\Pi$ = 132 atm

The van't Hoff's factor for glucose, $i$ = 1

$R = 0.08206 Latmmol^(-1)K^(-1)$

$T = 298 K$

Substituting the values in the above equation we get,

$132 atm = 1* M* 0.08206 Latmmol^(-1)K^(-1)* 298$

$M = (132 atm)/(1* 0.08206 Latmmol^(-1)K^(-1)* 298) = 5.4797 molL^(-1) \simeq 5.48 molL^(-1)$

So, the molarity of the solution is $5.48 molL^(-1)$ .

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If a drop of blood is 0.05 mL, how many drops of blood are in a blood collection tube that holds 8 mL ?

Answers

I think your answer is 20x

Final answer:

The total number of drops in an 8 mL blood collection tube is obtained by dividing the volume of the tube (8 mL) by the volume of one drop (0.05 mL), resulting in 160 drops.

Explanation:

To find out how many drops of blood are in the blood collection tube that holds 8mL, you have to divide the total volume of the tube by the volume of one drop. Since one drop of blood is 0.05mL, the calculation is as follows:

8 mL / 0.05 mL = 160 drops of blood

So, if a blood collection tube holds 8 mL, it would contain 160 drops of blood.

Learn more about Volume Calculation here:

brainly.com/question/32822827

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Be sure to answer all parts. Industrially, hydrogen gas can be prepared by combining propane gas (c3h8) with steam at about 400°c. The products are carbon monoxide (co) and hydrogen gas (h2). (a) write a balanced equation for the reaction. Include phase abbreviations. (b) how many kilograms of h2 can be obtained from 8.31 × 103 kg of propane

Answers

Balanced chemical reaction:

C₃H₈(g) + 3H₂O(g) → 3CO(g) + 7H₂(g).

M(C₃H₈) = 44.1 g/mol; molar mass of propane.

M(H₂) = 2 g/mol; molar mass of hydrogen.

From balanced chemical reaction: n(C₃H₈) : n(H₂) = 1 : 7.

7m(C₃H₈) : M(C₃H₈) = m(H₂) : M(H₂).

7·8310 kg : 44.1 g/mol = m(H₂) : 2 g/mol.

m(H₂) = 2638.09 kg; mass of hydrogen.

Answer: a) $C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $2.64* 10^3kg$

Explanation:

a) According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of propane}=(8.31* 10^6g)/(44.1g/mol)=0.188* 10^6moles$

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

According to stoichiometry:

1 mole of $C_3H_8$ gives 7 moles of $H_2$

Thus $0.188* 10^6moles$ moles of $C_3H_8$ will give = $(7)/(1)* 0.188* 10^6=1.32* 10^6moles$ of $H_2$

Mass of $H_2=moles* {\text {molar Mass}}=1.32* 10^6moles* 2g/mol=2.64* 10^6g=2.64* 10^3kg$

Thus $2.64* 10^3kg$ of $H_2$ can be obtained from $8.31* 10^3$ kg of propane

A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution? Use the periodic table in the toolbar if needed.

Answers

Answer : The mass of NaOH present in the solution is, 0.0625 grams

Explanation : Given,

Mass % = 25 %

Mass of solution = 0.250 g

Formula used :

$Mass\%=\frac{\text{Mass of}NaOH}{\text{Mass of solution}}* 100$

Now put all the given values in this formula, we get the mass of NaOH.

$25=\frac{\text{Mass of}NaOH}{0.250g}* 100$

$\text{Mass of}NaOH=0.0625g$

Therefore, the mass of NaOH present in the solution is, 0.0625 grams

The concentration of NaOH is 25.00% by mass, it means that 25.00% of the mass of the solution is of NaOH. Hence:

$m_(NaOH)=25.00\%* m_(solution)\Longrightarrow m_(NaOH)=(25)/(100)* 0.250~g\iff\n\n\boxed{m_(NaOH)=0.0625~g}$

All light travels at the same speed.
O True
O False

Answers

False

thank me later

Answer:

False

Explanation:

It depends on wixh kid of light is it

Example

Sun light difders from lamp light

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution? Kb = 0.51°C/m for water. Enter your answer with 2 decimal places and no units.

Answers

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_(solute)$ = Given mass of solute (CsCl) = 8.00 g

$M_(solute)$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_(solvent)$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2* 0.51^oC/m* (8.00* 1000)/(168.4g/mol* 92)\n\n\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?

Answers

Answer: 7.07 grams

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}*{\text{Molar Mass}}$

$\text{Moles of} zinc=(21g)/(65g/mol)=0.32moles$

$\text{Moles of} CuCl_2=(7g)/(134g/mol)=0.052moles$

$Zn+CuCl_2\rightarrow Cu+ZnCl_2$

According to stoichiometry :

1 mole of $CuCl_2$ require 1 mole of $Zn$

Thus 0.052 moles of $CuCl_2$ will require= $(1)/(1)* 0.052=0.052moles$ of $Zn$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Zn$ is the excess reagent.

As 1 mole of $CuCl_2$ give = 1 mole of $ZnCl_2$

Thus 0.052 moles of $CuCl_2$ give = $(1)/(1)* 0.052=0.052moles$ of $ZnCl_2$

Mass of $ZnCl_2=moles* {\text {Molar mass}}=0.052moles* 136g/mol=7.07g$

Thus 7.07 g of $ZnCl_2$ will be produced from the given masses of both reactants.

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