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2 years ago

How do the percent compositions for C3H6 and C4H7 compare?

Chemistry

1 answer:

mariarad [96]2 years ago

3 0

A. They are the same

<h3>Further explanation</h3>

Given

C3H6 and C4H8

Required

The percent compositions

Solution

C₃H₆(MW = 42 g/mol)

%C = 3.12/42 x 100% = 85.71%

%H = 6.1/42 x 1005 = 14.29%

C₄H₈(MW=56 g/mol)

%C = 4.12/56 x 100% = 85.71%

%H = 8.1/56 x 100%=14.29%

So they are the same, because mol ratio of C and H in both compounds is the same, 1: 2

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How many grams of oxygen gas are required to produce 65.75 grams of steam?

Nana76 [90]

Answer:

The mass of oxygen gas required to produce 65.75 grams of steam is approximately 162.2 grams

Explanation:

From the question, we have the following chemical reaction equation;

2C₃H₁₈(l) + 25O₂ (g) → 16CO₂(g) + 18H₂O (g)

The molar mass of oxygen, O₂ = 32 g/mol

The molar mass of steam, H₂O = 18.01528 g/mol

25 moles of oxygen are required to produce 18 moles of steam

Therefore, according to Proust's law of definite proportions;

(32 × 25) g of oxygen are required to produce (18 × 18.01528) g of steam

65.75 g of steam will be produced by (32 × 25)/(18 × 18.01528) × 65.75 g ≈ 162.2 g of oxygen O₂.

3 0

2 years ago

Plsss helppppppppp

Fantom [35]

Number one: True Number 2: False because hot stars apear as blue-white and cool stars apear as red

7 0

2 years ago

What is the volume of 500g of CO2?

Airida [17]

Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter

Try to see if this helps

3 0

3 years ago

Read 2 more answers

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

How much energy (heat) is required to convert 248 g of water from 0 oC to 154 oC? Assume that the water begins as a liquid, that

Nuetrik [128]

Answer:

The total heat required is 691,026.36 J

Explanation:

Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L

Where Q: amount of heat, m: mass and L: latent heat

On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).

In this case, the total heat required is calculated as:

Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C

Q= c*m*ΔT

$Q=4.184\frac{J}{g*C} *248 g* (100 -0 )C$

Q=103,763.2 J

Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then $\frac{248 g}{18 \frac{g}{mol} } =13.78 moles$ )

Q= m*L

$Q=13.78moles*40.79 \frac{kJ}{mol}$

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)

Q for temperature change from 100.0 ∘ C to 154 ∘ C, this is, the sensible heat of steam from 100 °C to 154°C.

Q= c*m*ΔT

$Q=1.99\frac{J}{g*C} *248 g* (154 - 100 )C$

Q=25,176.96 J

So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J

<u><em>The total heat required is 691,026.36 J</em></u>

8 0

3 years ago