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2 years ago

When the equation for the complete combustion of one mole of c3h7oh is balanced, the coefficient for oxygen is?

Chemistry

1 answer:

son4ous [18]2 years ago

3 0

C₃H₇OH + 4,5O₂ ----> 3CO₂ + 4H₂O

The coefficient for oxygen is 4,5.

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The energy transformation in an electromagnetic is from chemical to electrical to electromagnetic waves.

8_murik_8 [283]

True I think but I could be wrong

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2 years ago

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What is the empirical formula for C2H4(OH)2??

zhannawk [14.2K]

<span>Empirical formula for C2H4(OH)2 is</span><span>
C1H3O1

</span>

7 0

2 years ago

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A researcher studying the nutritional value of a new candy places a 4.70 g 4.70 g sample of the candy inside a bomb calorimeter

Sauron [17]

Answer : The nutritional Calories per gram of the candy are, 5.36 calories

Explanation :

First we have to calculate the amount of heat.

$q=c\times \Delta T$

where,

q = heat = ?

c = specific heat = $43.90kJ/K$

$\Delta T$ = change in temperature = $2.41^oC=2.41K$

Now put all the given values in the above formula, we get:

$q=43.90kJ/K\times 2.41K$

$q=105.799kJ$

Now we have to calculate the heat for per gram of sample.

Heat = $\frac{105.799kJ}{4.70g}=22.51kJ/g=22510J/g$

Now we have to calculate the heat in terms of calories.

As, 1 nutritional Calories = 1000 calories

and, 1 calories = 4.2 J

As, 4.2 J = 1 calories

So, 22510 J = $\frac{22510J}{4.2J}\times \frac{1}{1000}cal=5.36cal$

Therefore, the nutritional Calories per gram of the candy are, 5.36 calories

5 0

3 years ago

A compound contains 75% carbon and 25%

masha68 [24]

Answer:

CH4

Explanation:

Carbon: 75%/12grams = 6.25

6.25/6.25 = 1

Hyrdrogen: 25/1gram = 25

25/6.25 = 4

C1H4 = CH4

5 0

2 years ago

Calculate the pressure (psig) at the bottom of a 7500 foot column of fluid for each of the following fluids:

creativ13 [48]

Answer:

Pressure at the botton of 7500 ft column is:

a) P (water) = 3256.508 psig

b) P (drilling mud) = 4229.035 psig

c) P (gas lift) = 18.092 psig

d) P (Natural gas) = 17.092 psig

Explanation:

hidrostatic pressure:

P = ( d.f * g * h ) + Po

∴ density of fluid ( d.f ) ≡ Kg/m³

∴ aceleration o gravity ( g ) = 9.807 m/s²

∴ column height ( h ) = 7500 ft * ( m / 3.28084 ft ) = 2285.99 m

∴ atmospheric pressure (Po) = 101300 Pa ( Kg /m*s²)

specific gravity:

SG = d.f / d H2O (ref)

∴ d H2O = 997 Kg/m³

a) water:

⇒ P = ( 997 * 9.807 * 2285.99 ) + 101300

⇒ P = 22452844,88 Pa * ( psig / 6894.76 Pa ) = 3256.508 psig

b) Drilling Mud

⇒ SG = 1.3 = d.f / 997

⇒ d.f = 1296.1 Kg/m³

⇒ P = ( 1296.1 * 9.807 * 2285.99 ) + 101300

⇒ P = 29158182.16 Pa = 4229.035 psig

c) Gas lift column

⇒ SG = 0.85 = d.f / d.air (ref)

∴ d.air = 0.0765 Lb/ft³ = 1.23 Kg/m³

⇒ d.f = 1.23 * 0.85 = 1.0455 Kg/m³

⇒ P = ( 1.0455 * 9.807 * 2285.99 ) + 101300

⇒ P = 124738.755 Pa = 18.092 psig

d) Natural Gas column:

⇒ SG = 0.6 = d.f / 1.23

⇒ d.f = 0.738 Kg/m³

⇒ P = ( 0.738 * 9.807 * 2285.99 ) + 101300

⇒ P = 117845.003 Pa = 17.092 psig

4 0

2 years ago