All categories

2 years ago

A compound has a pH of 1 in solution, where it has completely lonized. The compound is a

Chemistry

1 answer:

Fantom [35]2 years ago

8 0

Answer:

3 no

Explanation:

Calculate pH and pOH of a weak acid or base solution ... acids and bases are completely ionized when dissolved in water. ... You have 50 mL acetic acid in 1 L vinegar.

You might be interested in

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago

The smallest particle to retain the properties of an element is an?

svlad2 [7]

I’m so sorry explain it more

4 0

2 years ago

What types of elements form covalent bonds?

spayn [35]

Covalent bonds form when electrons are shared between two nonmetals.

3 0

2 years ago

An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with

Kamila [148]

Answer:

The molar mass of the unknown acid is 89 g/mol

Explanation:

Step 1: The balanced equation

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

Step 2: Data given

Mass of the acid = 0.093 grams

volume = 250 mL

titrates with 0.16 M NaOH

adds 6.5 mL NaOH

Step 3: Calculate moles of NaOH

We know the concentration and volume of NaOH needed to neutralize the acid.

By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.

Moles = Molarity * Volume

Moles = 0.16 M * 0.0065 L

Moles = 0.00104 moles NaOH

Step 4: Calculate moles of the unknown acid:

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

For 0.00104 moles NaOH we have 0.00104 moles of HA

Step 5: Calculate the molar mass of the acid

Molar mass Ha = Mass Ha / moles Ha

Molar mass Ha = 0.093 grams / 0.00104 moles

Molar mass Ha = 89.42 g/mol ≈89 g/mol

The molar mass of the unknown acid is 89 g/mol

3 0

2 years ago

A teapot with a surface area of 785 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic

lina2011 [118]

Answer:

Given info: The surface area of teapot plated with silver is 700 cm2700 cm2, the cell is powered by 12.0-V12.0-V, the resistance of the cell is 1.80 Ω1.80 Ω, thickness of silver layer is 0.133-mm0.133-mm and density of silver is 10.5×103 kg/m310.5×103 kg/m3.

Write the expression for the mass of the silver.

m=ρAdm=ρAd (1)

Here,

ρρ is the density of silver.

AA is the surface area.

dd is the thickness of the silver layer.

Substitute 10.5×103 kg/m310.5×103 kg/m3 for ρρ, 700 cm2700 cm2 for AA and 0.133-mm0.133-mm for dd in equation (1) to find mass of the silver.

m=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 gm=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 g

Thus, the mass of silver is 97.8 g97.8 g.

Write the expression for number of moles of silver.

n=mWan=mWa (2)

Here,

mm is the mass of silver.

WaWa is the atomic weight.

The atomic weight of silver is 107.8 g/mole107.8 g/mole

Substitute 97.8 g97.8 g for mm, and 107.8 g/mole107.8 g/mole for WaWa in equation (2) to find number of moles of silver.

n=97.8 g107.8 g/mole=0.907 moln=97.8 g107.8 g/mole=0.907 mol

Thus, the number of moles of silver is 0.907 mol0.907 mol.

7 0

3 years ago