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3 years ago

The inert gas which is substituted for nitrogen in the air used by deep sea divers for breathing, is

Chemistry

1 answer:

Rudik [331]3 years ago

5 0

Answer:

1. C. helium

2. D. oxygen and acetylene

3. B. deliquescence

4. C. cutting very hard surfaces

5. B. 3.5%

Explanation:

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A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

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 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

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 = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

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

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

What is the term?

Alika [10]

Answer:

A DNA molecule

Explanation:

5 0

3 years ago

Help! Which of these is a compound?: Carbon, chlorine, uranium, ammonia.

MrRa [10]

Ammonia is the compound.

4 0

2 years ago

How many atoms of carbon are in a container that has 0.450 mol of CO2 and 2.55 mol of CaC2?

Fantom [35]

Answer:

The number of carbon atoms in the container is 1.806 × 10²⁴ or the container contains 1.806 × 10²⁴ atoms of carbon

Explanation:

By Avogadro's number, 1 mole of a substance contains 6.02 × 10²³ particles of the substance

Here we have 0.45 mole of CO₂ contains

0.45 × 6.02 × 10²³ particles of CO₂ that is 2.709 × 10²³ particles of CO₂ or equivalent to 2.709 × 10²³ atoms of Carbon

Similarly, 2.55 moles of CaC₂ contains 2.55 × 6.02 × 10²³ particles of CaC₂ or 1.5351 × 10²⁴ atoms of Carbon

The total number of carbon atoms is therefore;

2.709 × 10²³ + 1.5351 × 10²⁴ = 1.806 × 10²⁴ atoms of carbon.

5 0

3 years ago

Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?

TiliK225 [7]

Answer:- HBr is limiting reactant.

Solution:- The given balanced equation is:

$2Al+6HBr\rightarrow 2AlBr_3+3H_2$

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

$8molAl(\frac{6molHBr}{2molAl})$

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

7 0

2 years ago