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3 years ago

Write an equation to show how perchloric acid, HClO4, reacts with water. Include states of matter in your answer.

1 answer:

3 0

Since perchloric acid is a strong acid, it will dissociate completely, along with water. The products will be Perchlorate ions and Hydronium ions after the Perchloric acid gave away its proton and water, acting as a base, accepted that abandoned proton.

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HELPPP What do you think caused this land formation ?

Arturiano [62]

Answer:

Erosion/Weathering

Explanation:

Because the rock is exposed to the ocean, the constant push and pull of the water against the rock erodes the material away and moves it elsewhere. The wind likely aided in this process as well. This caused the big gap in the center of the rock.

6 0

3 years ago

Indicate whether the following are acidic, basic, or neutral solution

goldenfox [79]

You answer this by using the pH formula and and the relation of pH and pOH, pH = -log[H+] and 14 = pH + pOH. The correct classification are as follows:

A. [H2O+]=6.0x10^-12
basic

B. [H3O+]=1.4x10^-9
basic

C. [OH-]=5.0x10^-12
acidic

D. {OH-]=3.5x10^-10
acidic

Hope this answers the question.

5 0

3 years ago

If 8cm^3 of H2 reacts with an excess of Cl2, calculate how much of the HCL(g) is produced.

Rom4ik [11]

First, we need the balanced equation: H₂ + Cl₂ ---> 2HCl

since not much information is given, I am assuming we are at STP and that 22.4 Liters= 1 mol

1) let's convert the volume to moles using the molar volume of a gas. also we need to convert the cm₃ to mL, then to Liters.

8 cm³ (1 ml/ 1 cm³)(1 L/ 1000 mL) (1 mol/ 22.4 Liters)= 3.6x10⁻⁴ moles of H₂

2) let's use the mole ratio of the balanced equation to convert moles of H₂ to moles of HCl

3.6x10⁻⁴ mol H₂ (2 mol HCl/ 1 mol H₂)= 7.1x10⁻⁴ mol HCl

3) lastly, we convert the moles of HCl to grams using the molar mass.

molar mass of HCl= 1.01 + 35.5= 36.51 g/mol

7.1x10⁻⁴ mol HCl (36.51 g/mol)= 0.026 grams HCl

8 0

3 years ago

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

Step 1: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

Step 2: Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

Step 3: Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

3 years ago

Approximately how many formula units of NaCl are in 116.88g of table salt (NaCl), knowing that the molar mass of NaCl is 58.44g/

elena-14-01-66 [18.8K]

Answer:

116.88g of table salt (NaCl) contains two formula units

Explanation:

Now,

We know that 1 formula unit of sodium chloride has a molar mass of 58.44g/mol

Hence;

Mass of 1 formula unit = 58.44g

Mass of x formula units = 116.88g

x = 116.88g * 1 formula unit/58.44g

x = 2 formula units

Therefore;

116.88g of table salt (NaCl) contains two formula units

5 0

3 years ago

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