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4 years ago

If you pump air into cycle tyre a slight warming effect is noticed at valve stem why

1 answer:

6 0

As the air molecules move through the valve they have friction as they hit the walls, and its this friction that causes it to heat up.

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Determine the poh of 0.01 molar solution of carbonic acid

Veronika [31]

Hello!

Data:

Molar Mass of H2CO3 (carbonic acid)

H = 2*1 = 2 amu

C = 1*12 = 12 amu

O = 3*16 = 48 amu

------------------------

Molar Mass of H2CO3 = 2 + 12 + 48 = 62 g/mol

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:

M (molarity) = 0.01 M (Mol/L) → $1*10^{-2}\:M$

Use: Ka (ionization constant) = $4.4*10^{-7}$

$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$

$4.4*10^{-7} = 1*10^{-2}*\alpha^2$

$1*10^{-2}*\alpha^2 = 4.4*10^{-7}$

$\alpha^2 = \dfrac{4.4*10^{-7}}{1*10^{-2}}$

$\alpha^2 = 4.4*10^{-7-(-2)}$

$\alpha^2 = 4.4*10^{-7+2}$

$\alpha^2 = 4.4*10^{-5}$

$\alpha = \sqrt{4.4*10^{-5}}$

$\boxed{\alpha \approx 2.09*10^{-5}}$

Now, we will calculate the amount of Hydronium [H3O+] in carbonic acid (H2CO3), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$

$[ H_{3} O^+] = 1*10^{-2}* 2.09*10^{-5}$

$[ H_{3} O^+] = 2.09*10^{-2-5}$

$\boxed{[ H_{3} O^+] = 2.09*10^{-7}}$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:

$pH = \:?$

$[ H_{3} O^+] = 2.09*10^{-7}$

apply the data to formula

$pH = - log[H_{3} O^+]$

$pH = - log[2.09*10^{-7}]$

$pH = 7 - log\:2.09$

$pH = 7 - 0.32$

$\boxed{pH = 6.68}$

Note:. The pH <7, then we have an acidic solution (weak acid).

Now, let's find pOH by the following formula:

$pH + pOH = 14$

$6.68 + pOH = 14$

$pOH = 14 - 6.68$

$\boxed{\boxed{pOH = 7.32}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

3 0

3 years ago

Three samples of a solid substance composed of elements A and Z were prepared. The first contained 4.31 g A and 7.70 g Z. The se

murzikaleks [220]

Answer:

The percentage composition of the elements of the compound in the three samples is the same.

Explanation:

<em>The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements in the same proportion by mass.</em>

Sample A:

Mass of A = 4.31 g; mass of Z = 7.70 g

Total mass of sample = 12.01

Percentage mass of A in the sample = (4.31 * 100)/12.01 = 35.9 %

Percentage mass of Z in the sample = (7.70 * 100)/12.01 = 64.1 %

Sample B:

Percentage mass of A in the sample = 35.9 %

Percentage mass of Z in the sample = 64.1 %

Sample C:

Mass of A = 0.718 g; Total mass of sample = 2.00 g

mass of Z = mass of sample - mass of A = 2.00 g - 0.718 g = 1.282 g

Percentage mass of A in the sample = (0.718 * 100)/2.00 = 35.9 %

Percentage mass of Z in the sample = (1.282 * 100)/2.00 = 64.1 %

From the calculations, it can be seen that the percentage composition of the elements in the compound is the same for the three samples.

8 0

4 years ago

Sheila is on the track team and can run 500 yards in 40 seconds. How many meters can she run in 40 seconds?

tamaranim1 [39]

She can run 457.2 meters in 40 seconds

5 0

3 years ago

Why is it important to use scientific notation when talking about our solar system<br> and galaxies?

IgorLugansk [536]

Answer:

The main purpose for converting numbers into scientific notation is to make calculations with surpisingly large or small numbers less complicated. Since zeros are not used to place the decimal point, all of the digits in a number in scientific notation are meaningful and easier to read. Hope this helps!

Explanation:

3 0

3 years ago

According to the equation for this chemical reaction, what does the 2 in front of 2Ag tell us?

storchak [24]

The number of molecules involved in the reaction, in this case, the number of Ag atoms involved.

3 0

4 years ago

Read 2 more answers