All categories

3 years ago

Calculate the molarity of 0.700 mol of na2s in 1.05 l of solution.

1 answer:

6 0

Molarity is the amount of solute molecule (in moles) per 1L of solvent. In this case, the solute is 0.7mol Na2S and the solvent volume is 1.05L. Since the unit in this problem is already mol and L then you don't need to do any conversion of the units. The calculation would be:

molarity = mol of solute / (1L/ volume of solvent)
molarity = 0.7 mol/ (1L/ 1.05L)= 0.67M

You might be interested in

Estimate the air pressure at an altitude of 5km

r-ruslan [8.4K]

The answer is 100 kg

6 0

3 years ago

Classify the following aqueous solutions as: strong acid, weak acid, neutral, weak base, or strong base.

kirill [66]

Vinegar pH 3.2: Weak acid

Battery acid pH 0.5: Strong acid

Shampoo pH 7.0: Neutral

Ammonia pH 11.1 Strong base

4 0

1 year ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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 $CO_2=13.00g$

Mass of $H_2O=2.662g$

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 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

write an activity to show the change in state and change in temperature during a chemical reaction(change)

Paha777 [63]

If a piece of charcoal is taken and set on fire it will evolve CO2 and heat. Charcoal is solid but carbon dioxide is gaseous. This is an example of change of state and evolve of heat during chemical reaction.

5 0

3 years ago

An alkanol contains 37.5 carbon

kow [346]

Answer:

Empirical formula is CH₄

Molecular formula = C₂H₈

Explanation:

Mass of carbon = 37.5 g

Mass of hydrogen = 12.5 g

Molecular weight = 32 g/mol

Molecular formula = ?

Empirical formula = ?

Solution:

Number of gram atoms of C = 37.5 g /12g/mol = 3.125

Number of gram atoms of H = 12.5 g / 1.008 g/mol= 12.4

Atomic ratio:

C : H

3.125/3.125 : 12.4 /3.125

1 : 4

C : H : = 1 : 4

Empirical formula is CH₄

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 32 / 16

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 ( CH₄)

Molecular formula = C₂H₈

3 0

3 years ago