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2 years ago

Which of the following are functions of the nose?

Chemistry

2 answers:

aleksley [76]2 years ago

5 0

The second option would be the correct answer.

MArishka [77]2 years ago

3 0

Answer:

The answer should be I and III

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When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

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

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

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 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

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 = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

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

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

Which law explain why tire pressure increases on a summer day

Musya8 [376]

Boyle's law

I hope this helps

3 0

3 years ago

The reaction 2 ClO2(g) + F2(g) → 2 FClO2(g) is first-order in both ClO2 and F2. When the initial concentrations of ClO2 and F2 a

kupik [55]

Answer:

3. 75.0%

Explanation:

2 ClO2(g) + F2(g) → 2 FClO2(g)

First order with respect to ClO2 and F2.

This means the rate equation is given as;

Rate = k [ClO2][F2]

When the initial concentrations of ClO2 and F2 are equal?

Let's assume an initial value of 1 for both reactants, so rate equation is given as;

Rate = k * 1 * 1 = k

The rate after 25% of the F2 has reacted is what percent of the initial rate?

The concentration left of F2 is 75% ( 100% - 25%) = 0.75

Concentration of ClO2 remains 1.

So rate equation is given as;

Rate = k * 1 * 0.75 = 0.75 k

Comparing 0.75k and k.

This means our answer is;

3. 75.0%

6 0

3 years ago

Are we likely to run out of gold in the near future ?

sweet [91]

Earth isn’t likely going to be running out of gold for around 50-100 years

5 0

3 years ago

A recipe for chocolate cake calls for 236.6 mL of buttermilk.

Mice21 [21]

Answer:236.6

Explanation:

5 0

3 years ago