How do scientists measure the strength of acids and bases describe this scale?

Assuming you have 6.24 x 1014 electrons and the surface area of the pail is 0.2 m2, what is the charge density (C/m2)?

pentagon [3]

Answer:

σ = 4.998 E-4 C/m²

Explanation:

1 Coulomb (C) ≡ 6.241509 E18 electrons (e)

∴ # elect = 6.24 E14 elect

charge (Q):

⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C

charge density (σ):

σ = Q/S

∴ surface area (S) = 0.2 m²

⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)

⇒ σ = 4.998 E-4 C/m²

4 0

3 years ago

What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?

mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

$C=\frac{n}{V}$ .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is $K_{2}SO_{4}$ thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass

Putting the values,

$n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol$

Thus, number of moles of $K_{2}SO_{4}$ will be $0.1061\times 0.5=0.053 mol$ .

The volume of solution is 225 mL, converting this into L,

$1 mL=10^{-3}L$

Thus,

$225 mL=0.225 L$

Putting the values in equation (1),

$C=\frac{(0.053 mol}{0.225 L}=0.236 M$

Therefore, concentration of potassium sulfate solution is 0.236 M.

4 0

3 years ago

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

Answer: The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

3 years ago

Can a molecule with nonpolar bonds ever be polar? Why why not?

Tanya [424]

Answer:

A molecule that has only nonpolar bonds and no polar bonds cannot be polar.

Explanation:

However, a molecule that CONTAINS nonpolar bonds is different, because it can contain polar bonds. A molecule that contains nonpolar bonds can be polar as long as it also contains polar bonds.

Pls, choose me as brainliest!

5 0

2 years ago

Given 2.14g of B2H6

Shkiper50 [21]

Answer:

(A) is 0.0773 mol B2H6

(C) is 2.79 x 10^23 H atoms

Explanation:

Questions (A) and (B) are the same.

2.14 g B2H6 x (1 mol B2H6/27.668g B2H6) = 0.0773 mol B2H6 (A)

27.668 is the molar mass of B2H6 calculated from the period table:

(2 x 10.81) + (6 x 1.008) = 27.668

1.008 is the mass of H and 10.81 is the mass of B

(C)

0.0773 mol B2H6 x (6 mol H/ 1 mol B2H6) x (6.022 x 10^23 H atoms/1 mol H)

= 2.79 x 10^23 hydrogen atoms

Further Explanation:

For every 1 mol of B2H6, there are 6 moles of H (indicated by the subscript)

6.022 x 10^23 is Avogrado's number and it equals to 1 mol of anything

Avogrado's number can be in units of atoms, molecules, or particles

5 0

3 years ago