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2 years ago

What is the ph of a solution with [h3o+] = 1 × 10-9 m?

1 answer:

7 0

PH scale is used to determine how acidic, basic or neutral a solution is
pH can be calculated using the H₃O⁺
ph can be calculated as follows
pH = - log[ H₃O⁺]
[H₃O⁺] = 1 x 10⁻⁹
pH = - log [1 x 10⁻⁹]
pH = 9
pH of solution is 9

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According to Dalton, atoms cannot be SELECT ALL THAT APPLY a Created b Destroyed c Subdivided d Bonded

Sveta_85 [38]

Answer:

The answer is subdivided,created or destroyed

6 0

2 years ago

A chemist dissolves 867. mg of pure barium hydroxide in enough water to make up 170. mL of solution. Calculate the pH of the sol

Ilya [14]

Answer: The pH of the solution is 11.2

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Ba(OH)_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.867g}{171g/mol}=0.00507mol$ (1g=1000mg)

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.00507\times 1000}{170}$

$Molarity=0.0298$

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pOH=-\log [OH^-]$

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}$

According to stoichiometry,

1 mole of $Ba(OH)_2$ gives 2 mole of $OH^-$

Thus 0.0298 moles of $Ba(OH)_2$ gives = $\frac{2}{1}\times 0.0298=0.0596$ moles of $OH^-$

Putting in the values:

$pOH=-\log[0.0596]=2.82$

$pH+pOH=14$

$pH=14-2.82$

$pH=11.2$

Thus the pH of the solution is 11.2

8 0

3 years ago

calcium has atomic number of 20 how many electrons of a calcium atom are involved in chemical bonding

Lelu [443]

The answer is 2 electrons.

The electron configuration of calcium is 2:8:8:2

Calcium has two electrons in its outermost shell. These are its valence electrons and are the ones used in bonding with other elements. Valence electrons of an atom are those electrons that are in its outer energy shell or that are available for bonding.

Calcium is a metal. When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms forming ions. The resulting compound is known as an ionic compound.

For example, when calcium metal reacts with chlorine gas, calcium gives up its two valence electrons and Chlorine accepts them resulting in a new substance called calcium chloride in which the two elements have ended up forming ionic bonds.

7 0

3 years ago

Read 2 more answers

PLEASE HELP this is chemistry 12 points

Inessa05 [86]

Answer:

6.791

Explanation:

For proper significant figures with addition, you would use the significant figures of the number with lowest decimal place. 6.298 goes to the 10⁻³ place. 0.492712 goes to the 10⁻⁶ place. You will go out to the 10⁻³ place.

6.298 + 0.492712 = 6.790712 ≈ 6.791

8 0

3 years ago

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4 NH3 + 5 O2 → 4 NO + 6 H2O When 40.0 g NH3 and 50.0 g

luda_lava [24]

Answer:

18.75 g of NH3.

Explanation:

The balanced equation for the reaction is given below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.

This can be obtained as follow:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160 g

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Next, we shall determine the excess reactant. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.

From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.

Therefore, O2 is the limiting reactant and NH3 is the excess reactant.

Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, Xg of NH3 will react with 50 g of O2 i.e

Xg of NH3 = (68 × 50)/160

Xg of NH3 = 21.25 g

Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.

Finally, we shall determine mass of the remaining excess reactant as follow:

Mass of excess reactant = 40 g

Mass of excess reactant that reacted = 21.25 g

Mass of excess reactant remainig =?

Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)

Mass of excess reactant remainig

= 40 – 21.25

= 18.75 g

Therefore, the mass of excess reactant remaining is 18.75 g of NH3.

8 0

3 years ago