Answer : The reagent present in excess and remains unreacted is, 
Solution : Given,
Moles of 
= 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is,
Answer:
467
Explanation:
ncl2 = 454.4x1/(71.0 g/mol) = 6.40 mols cl2
6.40 mols cl2 x 2molsHCL/1moleCL2 x 36.5g/1moleHCL = <u>467 g HCL</u>
Answer: It's only one proton in the middle of the red circle there's nothing outside it.
Explanation: I'm not sure sorry :|
hope this helps
Answer:
Explanation:
formula of osmotic pressure is as follows
p= n RT
n is mole of solute per unit volume
If m be the grams of solute needed
m gram = m / 227.1 moles
m / 227.1 moles dissolved in .279 litres
n = m / (227.1 x .279 )
= m / 63.36
substituting the values in the osmotic pressure formula
5.14 = (m / 63.36) x .082 x 298
m / 63.36 = .21
m = 13.32 grams .
Answer:
<h3>The answer is 123.72 g</h3>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume = 157 mL
density = 0.788 g/mL
We have
mass = 0.788 × 157 = 123.716
We have the final answer as
<h3>123.72 g</h3>
Hope this helps you
