All categories

3 years ago

Calculate the volume of 4.00 molar NaOH solution required to prepare 100 mL of a 0.500 molar solution of NaOH.

Chemistry

1 answer:

Kazeer [188]3 years ago

7 0

Answer:

The answer to your question is V₁ = 12.5 ml

Explanation:

Data

Volume = V₁?

[NaOH] = C₁ = 4.0 M

Volume 2 = V₂ = 100 ml

[NaOH] = C₂ = 0.5 M

Formula of dilution

V₁C₁ = V₂C₂

Solve for V₁ (original solution)

V₁ = $\frac{V2C2}{C1}$

Substitution

V₁ = $\frac{(0.5)(100)}{4}$

Simplification

V₁ = $\frac{50}{4}$

Result

V₁ = 12.5 ml

You might be interested in

Lewis dot diagram for the Cs1+ ion

creativ13 [48]

Answer:

$Cs^+$

Explanation:

Cesium Lewis dot structure would look like this:

·Cs, because it only has one valence electron. But, since it has a plus, that means we lost an electron. So, we have to get rid of the dot and you have:

$Cs^+$

7 0

2 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant

schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

1 g = 1,000 mg
The molar mass of Cu is 63.55 g/mol
When 1 mole of Cu is deposited, 2 moles of electrons circulate.
The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
1 A = 1 C/s

The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

$336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s$