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:
Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).
The wavelengths of light that an atom gives off when an electron falls to a lower energy level corresponds to Emission spectrum , Option D is the correct answer.
<h3>What is Emission Spectrum ?</h3>
Light is absorbed or emitted when an electron jumps or falls into an energy level.
The energy of light absorbed or emitted is equal to the difference between the energy of the orbits.
Therefore , the wavelengths of light that an atom gives off when an electron falls to a lower energy level corresponds to Emission spectrum.
To know more about Emission Spectrum
brainly.com/question/13537021
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Answer:
%age Yield = 85.36 %
Solution:
The Balance Chemical Reaction is as follow,
C₆H₁₂O + Acid Catalyst → C₆H₁₀ + Acid Catalyst + H₂O
According to Equation ,
100 g (1 mole) C₆H₁₂O produces = 82 g (1 moles) of C₆H₁₀
So,
4.0 g of C₆H₁₂O will produce = X g of C₆H₁₀
Solving for X,
X = (4.0 g × 82 g) ÷ 100 g
X = 3.28 g of C₆H₁₀ (Theoretical Yield)
As we know,
%age Yield = (Actual Yield ÷ Theoretical Yield) × 100
%age Yield = (2.8 g ÷ 3.28 g) × 100
%age Yield = 85.36 %
Answer:
n = 7.86 mol
Explanation:
This question can be solved using the ideal gas law of PV = nRT.
Temperature must be in K, so we will convert 22.5C to 295 K ( Kelvin = C + 273).
R is the ideal gas constant of 0.0821.
(2.24atm)(85.0L) = n(0.0821)(295K)
Isolate n to get:
n = (2.24atm)(85.0L)/(0.0821)(295K)
n = 7.86 mol
