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3 years ago

Certain hydrocarbon is 92.3 % carbon and 7.7 % hydrogen by mass.If the molar mass of the hydrocarbon is approximately 40 g/mol,

Chemistry

1 answer:

zimovet [89]3 years ago

5 0

Answer:

The hydrocarbon has a molecular formula of C3H3

Explanation:

Step 1: Data given

Suppose the mass of the compound is 100 grams

A hydrocarbon contains:

⇒ 92.3 % carbon = 92.3 grams

⇒ 7.7 % hydrogen = 7.7 grams

Molar mass of the compound is 40 g/mol

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles carbon = 92.3 grams / 12.01 g/mol

Moles carbon = 7.69 moles

Moles hydrogen = 7.70 grams / 1.01 g/mol

Moles hydrogen = 7.62

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

Carbon: 7.69/7.62 = 1

Hydrogen: 7.62/7.62 = 1

The empirical formula is CH

The molar mass is 13.02 g/mol

Step 4: Calculate the molecular formula

n = 40 g/mol / 13.02 g/mol

n ≈ 3

We need to multiply the empirical formula by 3 to get the molecular formula

Molecular formula = 3*(CH) = C3H3

The hydrocarbon is C3H3

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A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea

jekas [21]

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely sensible heating. Hence, the heat released by the fuel is equal to the heat absorbed by the water because of principle of energy conservation. The heat released by the fuel is expressed by the following formula:

$Q = m\cdot c \cdot \Delta T$ (1)

Where:

$m$ - Mass of the sample, in grams.
$c$ - Specific heat of water, in calories per gram-degree Celsius.
$\Delta T$ - Temperature change, in degrees Celsius.

If we know that $m = 10\,g$ , $c = 1\,\frac{cal}{g\cdot ^{\circ}C}$ and $\Delta T = 9\,^{\circ}C$ , then the heat released by the fuel is:

$Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)$

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

7 0

2 years ago

Is cooking raw tomatoes a physical or a chemical change? Explain why?

salantis [7]

When you are asked a question like this, you can always ask yourself this question. Can I change it back after this change? For example, if you are burning wood, you cannot bring it back to wood after you burn it, therefore, it is a chemical change. However, if you boil and evaporate water, you can make the water condense again back into its liquid form. In this case, you cannot bring the tomato back to its raw state. Therefore, cooking raw tomatoes is a chemical change.

7 0

3 years ago

What mass of carbon dioxide is produced from the complete combustion of 5.30x10-3 g of methane?

pshichka [43]

Answer:

1.45 x 10⁻² g CO₂

Explanation:

To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Combustion of Methane:

1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

5.30 x 10⁻³ g CH₄ 1 mole 1 mole CO₂ 44.007 g
--------------------------- x ---------------- x --------------------- x ----------------- =
16.043 g 1 mole CH₄ 1 mole

= 0.0145 g CO₂

= 1.45 x 10⁻² g CO₂

8 0

2 years ago

Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$