15.3 litres of water will be produced if we take 1.7 litres of Hydrogen
Explanation:
Let's take a look over synthesis reaction;
<u> </u><u />
<u>Balancing the chemical reaction;</u>
<u> </u><u />
Thus, 2 moles of hydrogen molecules are required to form 2 moles of water molecules.
<u>Equating the molarity;</u>
<u /> =
(Since, the molecular mass of hyd and water is 2 and 18 respectively)
x=
x= 15.3 litres.
Thus,15.3 L of water will be produced if we take 1.7 litres of Hydrogen in a synthesis reaction.
Answer:
<h2>7.54 atm </h2>
Explanation:
The required pressure can be found by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
From the question we have
We have the final answer as
<h3>7.54 atm </h3>
Hope this helps you
Here we have to choose the specific iron ion preset in the compound FeO among the given ions.
The iron ion present in FeO is iron (II) ion.
The compound FeO is a neutral compound. Where the anion is oxide ion (O²⁻) thus the cation should also contain two unit positive charge. Thus Fe²⁺. Hence , the iron ion is Fe (II) ion.
The ferride ion means Fe²⁻ ion which cannot be present in FeO as the anion is already present which is O²⁻.
The iron (I) ion means iron present in +1 state, then to neutralize the compound 2 iron (I) is needed but in compound FeO there is only one iron present. Thus it cannot be iron (I) ion.
The ferride (I) ion means Fe¹⁻, which cannot be the state of the iron in FeO as there already present an anion which is O²⁻ ion.
Hey There !:
Molar mass of H₂O = 18.01 g/mol
Therefore:
1 mol H₂O ----------------- 18.01 g/mol
9.00 moles ---------------- ( mass of H₂O )
mass of H₂O = 9.00 x 18.01 / 1
mass of H₂O = 162.09 / 1
mass of H₂O = 162.09 grams
I hope that it helps !
Answer:
12.33 cal/sec
Explanation:
As we know,
1 Kcal = 1000 cal
So,
0.74 Kcal = X cal
Solving for X,
X = (0.74 Kcal × 1000 cal) ÷ 1 Kcal
X = 740 cal
Also we know that,
1 Minute = 60 Seconds
Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,
= 740 cal / 60 sec
= 12.33 cal/sec