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PtichkaEL [24]
3 years ago
15

How would you write the conversion reaction from oil to biodiesel

Chemistry
1 answer:
Rashid [163]3 years ago
7 0

Answer:

Oil + alcohol -> glycerin

Explanation:

hope that this is what you meant

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a weather balloon contains 8.80 moles of helium at a ppressure of 0.992 atm and a temperature os 25 C at ground level. What is t
djverab [1.8K]
We can use the ideal gas law equation to find the volume of the balloon.
PV = nRT 
where 
P - pressure - 0.992 atm x 101 325 Pa/atm = 100 514 Pa
V - volume 
n - number of moles - 8.80 mol 
R - universal gas constant  - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 25 °C + 273 = 298 K
Substituting these values in the equation 
100 514 Pa x V = 8.80 mol x  8.314 Jmol⁻¹K⁻¹ x 298 K
V = 217 L
volume of balloon is 217 L
7 0
3 years ago
What are fossils?
siniylev [52]

answer: C

Explanation:

They are remains that are in rock

5 0
4 years ago
Read 2 more answers
Find the volume occupied by 4.00 g of helium gas at <br> 25.0o C and 1.18 atm of pressure.
Triss [41]

Answer: 41.46 L

Explanation:

La ecuación que describe  relación entre presión, volumen,  temperatura y la cantidad (en moles)

de un gas ideal es:   PV = nRT

Donde:  P = Presión absoluta ,  V= Volumen  , n = Moles de gas , R = Constante universal de los gases ideales,  T = Temperatura absoluta,

R = 0.082 L. atm/mol. °K

V = nRT/P

Calculanting n

n = mass/ molecular mass

<h3>n = 4 g / 2g. mol⁻¹</h3><h3>n = 2 mol</h3><h3>T =25⁰ + 273 ⁰K = 298 ⁰K</h3><h3>V = (2 mol ₓ0.082 L. atm / mol.°K x 298 ⁰K) / 1.18 atm = 41.46 L</h3>
6 0
3 years ago
In order to perform a chemical reaction, 225 mL of 0.500 M lead (II) nitrate is required. How much 5.00 M stock solution is requ
vladimir1956 [14]

Answer:

what

Explanation:

what ````

4 0
3 years ago
Consider the reaction: 2NO(g) O2(g) --&gt; 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is
poizon [28]

Answer:

a. 0.0022 M/s

b. 0.0011 M/s

Explanation:

Let's consider the following reaction.

2 NO(g) + O₂(g) → 2 NO₂(g)

The rate of disappearance of NO is 0.0022 mol NO. L⁻¹.s⁻¹

<em>a. At what rate is NO₂ being formed?</em>

The molar ratio of NO to NO₂ is 2:2. The rate of formation of NO₂ is:

\frac{0.0022molNO}{L.s} .\frac{2molNO_{2}}{2molNO} =\frac{0.0022molNO_{2}}{L.s}

<em>b. At what rate is molecular oxygen reacting?</em>

The molar ratio of NO to O₂ is 2:1. The rate of disappearance of O₂ is:

\frac{0.0022molNO}{L.s} .\frac{1molO_{2}}{2molNO} =\frac{0.0011molO_{2}}{L.s}

8 0
3 years ago
Read 2 more answers
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