Answer:
Ammonium chloride has ammonium (NH4 +) part which is slightly acidic and Barium Hydroxide is a strong base.
You can see that an acid-base reaction will occur.
In this case :
Its highly endothermic reaction
Forms aqueous ammonia (mind that ammonia is highly water soluble gas so no effervesence) and Barium Chloride which is water soluble.
Ba(OH)2 +NH4Cl ——->BaCl2 + NH3 +H2O (unbalanced)
Answer:
The moles of sucrose that are available for this reaction is 0.0292 moles
Explanation:
Combustion is an specifyc reaction where the reactants react with O₂ in order to produce CO₂ and H₂O
This combustion is: C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
We have to conver the mass to moles, to find out the limiting reactant
10 g . 1 mol / 342 g = 0.0292 moles of sucrose
8 g . 1mol / 32g = 0.250 moles of O₂
The moles of sucrose that are available for this reaction is 0.0292 moles
Before we start to work with the equation we must find the limiting reactant. When you find it, you can do all the calculations.
Answer:
For this angular momentum, no quantum number exist
Explanation:
From the question we are told that
The magnitude of the angular momentum is 
The generally formula for Orbital angular momentum is mathematically represented as
Where 
is the quantum number
now
We can look at the given angular momentum in this form as
comparing this equation to the generally equation for Orbital angular momentum
We see that there is no quantum number that would satisfy this equation
13.4 billion years is 3 times of the half-life, 4.47 billion years. So the Uranium-238 will go through three times of half decay. So the remain percentage will be 50%*50%*50%=12.5%.
Answer:
The answer will be 2.98K
Explanation:
Using the formula:
Q = mc∆T
Q= 5,800 (heat in joules)
m= convert 15.2kg to g which is 15200g (mass in grams)
c= 0.128 J/g °c (Specific heat capacity)
∆T= what we need to find (temperature change)
5800J = 15200g x 0.128 x ∆T
= 2.98K
