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2 years ago

What is the density of 53.4 wt aqueous naoh?

1 answer:

4 0

Missing question: what is the density of 53.4 wt% aqueous NaOH if 16.7 mL of the solution diluted to 2.00L gives 0.169 M NaOH?

Answer is: density is 1.52 g/mL.

c₁(NaOH) = ?; molarity of concentrated sodium hydroxide.

V₁(NaOH) = 16.7 mL; volume of concentrated sodium hydroxide.

c₂(NaOH) = 0.169 M; molarity of diluted sodium hydroxide.

V₂(NaOH) = 2.00 L · 1000 mL/L = 2000 mL; volume of diluted sodium hydroxide.

Use equation: c₁V₁ = c₂V₂.

c₁ = c₂V₂ / V₁.

c₁ = 0.169 M · 2000 mL / 16.7 mL.

c₁(NaOH) = 20.23 M.

m(NaOH) = 20.23 mol · 40 g/ml.

m(NaOH) = 809.53 g.

The mass fraction is the ratio of one substance (in this example sodium hydroxide) with mass to the mass of the total mixture (solution).

Make proportion: m(NaOH) : m(solution) = 53.4 g : 100 g.

m(solution) = 1516 g in one liter of solution.

d(solution) = 1516 g/L = 1.52 g/mL.

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You wish to make a buffer with pH 7.0. You combine 0.060 grams of acetic acid and 14.59 grams of sodium acetate and add water to

aleksandr82 [10.1K]

Answer:

The pH of the buffer is 7.0 and this pH is not useful to pH 7.0

Explanation:

The pH of a buffer is obtained by using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pH is the pH of the buffer

The pKa of acetic acid is 4.74.

[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles

[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles

Replacing:

pH = 4.74 + log [0.1779mol] / [0.001mol]

pH = 6.99 ≈ 7.0

The pH of the buffer is 7.0

But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,

this pH is not useful to pH 7.0

7 0

2 years ago

An atom in the ground state contains a total of 5 electrons 5 protons and 5 neutrons. Which Lewis electron-dot diagram represent

Luba_88 [7]

I don't know if you didn't gave a picture choice or if i didn't get the picture.
But lets call this atom A. Electron dot formula doesn't require Neutron and Protons, its main concern is valance elections.

So atom A has 5 electrons which means 2,3 it has 3 valance electrons. Its dot formula will become
:A.

I hope this helped.

7 0

2 years ago

A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not

AleksAgata [21]

Volume of tank = $6.9 m^{3}$ (given)

Since, $1 m = 3.28 ft$

So,

$1 m^{3} = 35.287 ft^{3}$

For $6.9 m^{3}$ :

$6.9\times 35.287 ft^{3} = 243.4803 ft^{3}$

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

$243.4803 ft^{3} \simeq 243.5 ft^{3}$

Hence, volume in $ft^{3}$ is 243.5.

5 1

2 years ago

Read 2 more answers

g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole

Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

3 0

2 years ago

Groups of atoms that are added to carbon backbones and give them unique properties are known as

Irina-Kira [14]

Answer:

Groups of atoms that are added to carbon backbones and give them unique properties are known as Functional Groups.

Explanation:

In organic chemistry they are called as Functional Group because they are the active part of a molecule. These groups give a unique characteristic to molecule both chemically and physically. Also, each functional group represent a different class of compounds.

Examples:

S No. Functional Group Name

1 R--X Alkyl Halides

2 R--OH Alcohols

3 R--NH₂ Amines

4 R--O--R Ethers

5 R--CO--R Ketones

6 R--CO--H Aldehydes

7 R--CO--OH Carboxylic acids

8 R--CO--X Acid Halides

10 R--CO--NR₂ Acid Amides

11 R--CO-OR' Esters

3 0

2 years ago