The correct reaction equation is:
Answer:
b) 1 mole of water is produced for every mole of carbon dioxide produced.
Explanation: <u>CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.</u>
a) <u>22.4 L of 
gas</u> is produced only when <u>
L of 
</u> is reacted with 22.4 L of 
. So it is wrong.
b) Since in the chemical equation the stoichiometric coefficient of 
and 
are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. <u>Therefore it is correct option.</u>
c) 
molecules is equal 1 mole of 
if produced then 3 moles of is required, which is not given in the option. So it is wrong.
d) 54 g of water or 3 moles of (<em>Molecular Weight of water is 18 g</em>) is produced when 3 moles of is used but in this option only one mole of is given. So it is wrong.
2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)
First off.. not a chem board.. but n e way.
This is a limiting reagent problem.
set it up as a DA problem.(Dimension Analysis)
Start with what you want.
you want Grams of acrylonitrile (C3H3N)
so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)
(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)
solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g
Same setup for the two other reactants.
so i did it and for
oxygen I got 11.04 grams
and for Ammonia i got 15.29 grams
So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.
Both the other reactants are in excess.
rate brainliest pls
Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
Answer:
See explanation below
Explanation:
What we have to consider is the hybridation of the three carbon atoms we are asked in this question .
Hybridization # bonds Angle
sp³ 4 109.5º
sp² 3 + 1 pi bond 120º
sp 2 + 2 pi bonds 180º
Carbon atom (a) is bonded to two atoms: Carbon (b) and one Hydrogen. It has a triple bond to Carbon (b). Therefore its hybridization is sp with two pi bonds, and for sp hybridization we know the angle is 180 º.
The same hybridization sp happens to carbon (b) bonded to Carbon (c) and C(a) using one sp bond to Carbon (a) and 2 pi bonds; it is also bonded using the other sp to Carbon (c). The angle is therefore 180 between Carbons b and c.
Carbon C is bonded to 4 atoms, therefore, its hybridization is sp³ and the angles with these 4 atoms will be 109.5 º tehedral ( one bond to OH, one to C(b), and 2 to H ) .
Question:
Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.
A. 0.272 mol
B. 0.04 mol
C. 0.997 mol
D. 0.139 mol
E. 0.0681 mol
Answer:
The correct option is;
E. 0.0681 mol
Explanation:
The equation for the reaction is
Zn + HCl = H₂ + ZnCl₂
Vapor pressure of the liquid = 18 torr = 2399.803 Pa
Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm
= 100007.775 Pa
Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation
Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa
Volume of H₂ = 1.7 L = 0.0017 m³
Temperature = 20 °C = 293.15 K
Therefore,
Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.
