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2 years ago

Calculate the pH of a solution that is 0.113 M HCI.

Chemistry

1 answer:

Alex Ar [27]2 years ago

6 0

Simplified concentration

$\\ \rm\Rrightarrow 1.13\times 10^{-2}M$

Now

$\\ \rm\Rrightarrow pH=-log[H^+]$

$\\ \rm\Rrightarrow pH=-log(1.13\times 10^{-2})$

$\\ \rm\Rrightarrow pH=-log1.13-log10^{-2}$

$\\ \rm\Rrightarrow pH=-0.05+2$

$\\ \rm\Rrightarrow pH=1.95$

Highly acidic

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2) Calculate the mass(g) of the following substances: b) 0.119 mole MgCl2

VARVARA [1.3K]

Answer:

MgCl2 = 24 + 2(35.5)

= 95

mass of substance = mol × molar mass

= 0.119 × 95

= 11.305 g

4 0

3 years ago

The element Oxygen, O, is an example of which of the following?

Anon25 [30]

It is an example of a molecule

5 0

3 years ago

Read 2 more answers

If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai

Rama09 [41]

We are given –

Final velocity of car is, v= 0
Initial velocity of car is, u= 100 km/hr
Time taken, t is = 3 minutes or 180 sec

Here–

$\qquad$ $\pink{\bf \longrightarrow Initial\: velocity = 100 \:km/hr}$

$\qquad$ $\sf \longrightarrow Initial\: velocity = \dfrac{ 100 \times 1000}{3600} \:m/s$

$\qquad$ $\pink{ \bf \longrightarrow Initial\: velocity = 27.78\: m/s}$

Now –

$\qquad$ ____________________________

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{Final\: Velocity -Initial \:Velocity }{Time}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{v -u}{t}}$

$\qquad$ $\sf \longrightarrow Acceleration = \dfrac{(0- 27.78)}{1800}$

$\qquad$ $\sf \longrightarrow Acceleration =\cancel{ \dfrac{- 27.78}{1800}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = -0.15 \: m/s^2}$

$\qquad$ _______________________________

6 0

2 years ago

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 13. g of butane is m

12345 [234]

Answer: The maximum amount of water that could be produced by the chemical reaction is 20.16 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For butane:

Given mass of butane = 13 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of butane}=\frac{13g}{58.12g/mol}=0.224mol$

For oxygen gas:

Given mass of oxygen gas = 70.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{70.9g}{32g/mol}=2.216mol$

The chemical equation for the reaction of butane and oxygen gas follows:

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

By Stoichiometry of the reaction:

2 moles of butane reacts with 13 moles of oxygen gas

So, 0.224 moles of butane will react with = $\frac{13}{2}\times 0.224=1.456mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, butane is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water

So, 0.224 moles of butane will produce = $\frac{10}{2}\times 0.224=1.12moles$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.12 moles

Putting values in equation 1, we get:

$1.12mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.12mol\times 18g/mol)=20.16g$

Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams

8 0

3 years ago

A glass holds 6 ounces of 60-proof rum (30 alcohol). how much fruit juice must be added to the rum so that the mixture is dilute

Elenna [48]

We need an equation that will relate the concentrated mixture and the diluted one. To solve this we use the equation,

M1 V1 = M2 V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

30 % x 6 oz = 20 % x V2

V2 = 9 oz

The volume of the diluted mixture would be 9 oz. Therefore, you will need to add 9 oz - 6 oz = 3 oz of fruit juice to dilute the 30 percent alcohol to 20 percent alcohol.

7 0

3 years ago