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Mandarinka [93]
4 years ago
15

Calculate the concentration of H3O⁺ in a solution that contains 5.5 × 10-5 M OH⁻ at 25°C. Identify the solution as acidic, basic

, or neutral. A) 1.8 × 10-10 M, basic B) 1.8 × 10-10 M, acidic C) 5.5 × 10-10 M, neutral D) 9.2 × 10-1 M, acidic E) 9.2 × 10-1 M, basic
Chemistry
1 answer:
harina [27]4 years ago
3 0

Answer : The correct option is, (A) 1.8\times 10^{-10}M, basic.

Explanation : Given,

Concentration of OH^- ion = 5.5\times 10^{-5}M

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (5.5\times 10^{-5})

pOH=4.26

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.26=9.74

Now we have to calculate the H_3O^+ concentration.

pH=-\log [H_3O^+]

9.74=-\log [H_3O^+]

[H_3O^+]=1.8\times 10^{-10}M

As we know that, when the pH value is less than 7 then the solution acidic in nature and when the pH value is more than 7 then the solution basic in nature.

From the pH value, 9.74 we conclude that the solution is basic in nature because the value of pH is greater than 7.

Therefore, the H_3O^+ concentration is, 1.8\times 10^{-10}M, basic.

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Density can be calculated using the following rule:
density = mass / volume
Therefore,
volume = mass / density

For diamond:
we have mass = 0.5 grams and density = 3.51 g / cm^3
Substituting in the rule, we can calculate the volume of diamond as follows:
volume = 0.5 / 3.51 = 0.14245 cm^3

For graphite:
we have mass = 0.5 grams and density = 2.25 g / cm^3
Substituting in the rule, we can calculate the volume of graphite as follows:
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Which statement compares the relative strength of the electric forces that hold particles together
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3 years ago
A sample of gas is held at 1000C at a volume of 20 L. If the volume is increased to 40 L, what is the new temperature of the gas
kaheart [24]

Answer:

The new temperature will be 2546 K or 2273 °C

Explanation:

Step 1: Data given

The initial temperature = 1000 °C =1273 K

The volume = 20L

The volume increases to 40 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 20L

⇒with T1 = the initial temperature = 1273 K

⇒with V2 = the increased volume = 40L

⇒with T2 = the new temperature = TO BE DETERMINED

20L/ 1273 K = 40L / T2

T2 = 40L / (20L/1273K)

T2 = 2546 K

The new temperature will be 2546 K

This is 2546-273 = 2273 °C

Since the volume is doubled, the temperature is doubled as well

8 0
3 years ago
What mass of salt would you need to add to 1.00kg of water to achieve a freezing point of -5 degrees C
Serga [27]

Answer:

The type of salt to be added to the water is not known from the question but no worries, I will try to give you the step by step procedure to answer any type of question similar to this.

To answer this question, we should know some facts.

1. the molar freezing point depression constant of water (Kf) = 1.86 K kg/mol

2. the molar mass of the salt if NaCl = 58.5 g/mol ; KCl = 74.5 g/mol

3. since the salt can dissociate if NaCl or KCl into two ions, the Van't Hoff factor ( i )= 2

Note that: the change in freezing point, molarity, deepression constant and van't Hoff factor are related by this formula;

                ΔTf = i Kf m

So lets take NaCl as the salt:

Molar mass = 58.5 g/mol

Van't Hoff factor = 2

1. calculate the number of moles

So we can calculate the molarity of the salt NaCl from the formula;

m = ΔTf / i Kf

m = 5 / 2 * 1.86

m = 5 / 3.72

m = 1.344 mol/kg

2. calculate the number of moles of the salt required

Next is to multiply the molarity by the mass of water. Density of water = 1kg/L

number of moles = 1.344 mol/kg * 1 Kg/L * 1 kg water

number of moles = 1.344 moles.

3. calculate the mass of the salt.

numner of moles = mass / molar mass

mass = number of moles * molar mass

mass = 1.344 * 58.5

mass = 78.624 g of NaCl salt.

You can follow these steps to solve for the type of salt you are given in the question.

6 0
4 years ago
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