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3 years ago

What is the difference between hypertonic and hypotonic

1 answer:

5 0

Answer: A solution will be hypertonic to a cell if its solute concentration is higher than that inside the cell, and the solutes cannot cross the membrane. If a cell is placed in a hypotonic solution, there will be a net flow of water into the cell, and the cell will gain volume.

Picture:
Meaning of hypertonic: (1): having a higher osmotic pressure than a particular fluid, typically a body fluid or intracellular fluid. (2) of or in a state of abnormally high muscle tone.

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A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

you have to prepare a ph 5.03 buffer, and you have the following 0.10m solutions available: hcooh, hcoona, ch3cooh, ch3coona, hc

TiliK225 [7]

Solution :

The buffer is the one that contains weak acid (the $\text{pK}_a$ is nearly equal to the required pH) and the salt of its conjugate base.

The $\text{pK}_a$ values of the given weak acids are as follows :

HCOOH : $\text{pK}_a$ = 3.744

$CH_3COOH : pK_a = 4.756$

HCN : $\text{pK}_a$ = 9.21

Since the required pH is 5.03, the suitable buffer is the mixture of the acetic acid and the salt of its conjugate base.

Let suppose the volumes of $CH_3COOH$ and $CH_3COONa$ are x and y mL respectively.

So the total volume of the buffer is 1000 mL.

∴ x+ y = 1000 ................(1)

Writing the Henderson-Hasselbalch equation for the given buffer solution :

$pH = pK_a + \log \ \frac{[CH_3COONa]}{[CH_3COOH]}$ .............(2)

$5.03 = 4.756 + \log \ \frac{\left(\frac{y \ mL \times 0.10 \ M}{(x+y) \ mL}\right)}{\left(\frac{x \ mL \times 0.10 \ M}{(x+y) \ mL}\right)}$

$5.03 = 4.756+ \log \frac{y}{x}$

$\frac{y}{x}=10^{5.03-4.756}$

y = 1.9 x

Substituting the values of y in equation (1), we get

x + 1.9 x = 1000

x = 345

Putting the value of x in (1), we get

345 + y = 1000

y = 655

Therefore the volume of $CH_3COOH$ is 345 mL and the volume of $CH_3COONa$ is 655 mL.

5 0

3 years ago

Which of the following atoms has the largest radius?<br> Mg<br> Si<br> S<br> Cl

timofeeve [1]

Answer:

Hope this helps :D sorry if im wrong :(

8 0

3 years ago

How many milliliters of 0.500 m kcl solution would you need to dilute to make 100.0 ml of 0.100 m kcl?

vagabundo [1.1K]

0.5XV=100X0.1
=> V=20ml

5 0

3 years ago

Why do we call meteors shooting stars?

zlopas [31]

They burn up and cause it to look like shooting stars

6 0

4 years ago