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Leni [432]
3 years ago
7

A 0.250 g sample of hydrocarbon (containing only carbon and hydrogen) undergoes complete combustion to produce 0.845 g of CO2 an

d 0.173 g of H2O. What is the empirical formula of this compound?
Chemistry
1 answer:
melomori [17]3 years ago
5 0

Answer:

CH

Explanation:

We have to obtain the mass of carbon and hydrogen in CO2 and H2O respectively.

For carbon in CO2;

0.845 * 12/44 = 0.23 g

For hydrogen in H20;

0.173 * 2/18 = 0.019 g

We convert the masses to moles of carbon and hydrogen

For carbon - 0.23/ 12 = 0.019 moles

For hydrogen - 0.019/1 = 0.019 moles

Dividing by 0.019 moles for carbon and hydrogen we have the emperical formula of the compound as CH

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In general for main-group elements, atomic size _____ from left to right across a period and _____ down a group.
liberstina [14]

Answer:

decreases, increases

Explanation:

In general for main-group elements atomic size decreases from left to right across a period and increases down a group

8 0
1 year ago
A reaction has ∆H = −356 kJ and ∆S = −36 J/K. Calculate ∆G (kJ) at 25°C.
Cloud [144]

Answer: -345.2 KJ

Explanation: As we know that ,dG=dH-TdS

T=25+273=298 K

dG= -356 x1000-298(-36)= -356000+10728

=-345272 j

= -345.2 KJ

5 0
2 years ago
What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

7 0
2 years ago
12. How is the melting point of a substance related to its freezing point? Both indicate the temperature at which the solid and
krok68 [10]
Both indicate the temperature at which the solid and liquid states of a substance are in equilibrium would be your answer.

This is beacause the melting point of a substance is the same as the freezing point of a substance. At this particular temp, the substance can be either a solid or a liquid. 

hope this helps!
8 0
3 years ago
Read 2 more answers
Explain why the replacement of the glutamic acid by valine changes the way that molecules of b-globin interact with each other.
aleksandr82 [10.1K]
For the answer to the question above, <span>Hydrophobic regions and hydrophilic regions in the molecules of the b-globin. The replacement causes these hemoglobin molecules to be stickies which gives the cell its sickle shape.
I hope this helps. Have a nice day!</span>
3 0
3 years ago
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