<h3><u>Answer;</u></h3>
NH3/NH4+
<h3><u>Explanation;</u></h3>
From the equation;
NH3(aq)+HNO3(aq)→NH4+(aq)+NO3−(aq)
NH3 is the base; while NH4+ is the conjugate acid
HNO3 is the acid; while NO3- is the conjugate base
- The conjugate base of a Brønsted-Lowry acid is species that is formed after an acid donates a proton while the conjugate acid of a Brønsted-Lowry base is the species formed after a base accepts a proton.
Answer:
False
Explanation:
Heres what its made of:
The core is made of hot, dense plasma (ions and electrons), at a pressure estimated at 265 billion bar (3.84 trillion psi or 26.5 petapascals (PPa)) at the center. Due to fusion, the composition of the solar plasma drops from 68–70% hydrogen by mass at the outer core, to 34% hydrogen at the core/Sun center.
Answer:The environmental problems directly related to energy production and Electricity is a secondary energy source that is generated producedfrom primary energy sources:
hope that helps im new so have a good day (:
Explanation:
Answer:
C. 100.7 amu
Explanation:
Isotopes of an element are atoms of an element with the same atomic number but different atomic masses. Each atomic mass of an isotope is known as an isotopic mass. An element that exhibits isotope, that is, that have two or more isotopes has a relative atomic mass that is not a whole number.
Relative atomic mass of X is the sum of the products of the relative abundances of each isotope and its isotopic mass.
For Isotope ¹⁰⁰X: 30% × 100 = 30 amu
For Isotope ¹⁰¹X: 70% × 101 = 70.7 amu
Relative atomic mass of X = (30 + 70.7) amu = 100.7 amu
Therefore, the approximate atomic mass of X is 100.7 amu
Answer:
Step 1;
q = w = -0.52571 kJ, ΔS = 0.876 J/K
Step 2
q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ
Explanation:
The given parameters are;
= 100 N·m
= 327 K
= 90 N·m
Step 1
For isothermal expansion, we have;
ΔU = ΔH = 0
w = n·R·T·ln(/) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71
w ≈<em> -0.52571</em> kJ
At state 1, q = w = -0.52571 kJ
ΔS = -n·R·ln(/) = -1 × 8.314 × ln(90/100) ≈ 0.876
ΔS ≈ 0.876 J/K
Step 2
q = 0 for adiabatic process
ΔU = 25×(27 - 327) = -7,500
w = ΔU = <em>-7.5 kJ</em>
ΔH = ΔU + n·R·ΔT
ΔH = -7,500 + 8.3142 × 300 = -5,005.74
ΔH = ΔU = <em>-5.00574 kJ</em>
