Answer: nitrogen
Explanation:
- Converting 5.78 g of hydrogen to moles, we know that the formula mass of hydrogen is about 2(1.00794)=2.01588 g/mol, so 5.78 grams is about 5.78/2.01588=2.867 mol.
- Converting 6.28 g of nitrogen to moles, we know that the formula mass of nitrogen is about 2(14.0067)=28.0134 g/mol, so 6.28 grams is about 6.28/28.0134 = 0.22417 mol.
From the equation, we know that for every 3 moles of hydrogen consumed, 1 mole of nitrogen is consumed.
- Considering the hydrogen, the reaction can occur 2.867/3=0.955 times.
- Considering the nitrogen, the reaction can occur 0.22417 times.
Therefore, <u>nitrogen</u> is the limiting reactant.
I believe it is aluminum
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1. <span>EACH ORBITS HOLDS A FIXED NUMBER OF ELECTRONS </span>
Answer:
75 kJ/mol
Explanation:
The reactions occur at a rate, which means that the concentration of the reagents decays at a time. The rate law is a function of the concentrations and of the rate constant (k) which depends on the temperature of the reaction.
The activation energy (Ea) is the minimum energy that the reagents must have so the reaction will happen. The rate constant is related to the activation energy by the Arrhenius equation:
ln(k) = ln(A) -Ea/RT
Where A is a constant of the reaction, which doesn't depend on the temperature, R is the gas constant (8.314 J/mol.K), and T is the temperature. So, for two different temperatures, if we make the difference between the two equations:
ln(k1) - ln(k2) = ln(A) - Ea/RT1 - ln(A) + Ea/RT2
ln (k1/k2) = (Ea/R)*(1/T2 - 1/T1)
k1 = 8.3x10⁸, T1 = 142.0°C = 415 K
k2 = 6.9x10⁶, T2 = 67.0°C = 340 K
ln(8.3x10⁸/6.9x10⁶) = (Ea/8.314)*(1/340 - 1/415)
4.8 = 6.39x10⁻⁵Ea
Ea = 75078 J/mol
Ea = 75 kJ/mol
Answer:
Explanation:
Let A₀ = the original amount of ⁵⁵Co
.
The amount remaining after one half-life is ½A₀.
After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.
After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.
The general formula for the amount remaining is:
A =A₀(½)ⁿ
where n is the number of half-lives
n = t/t_½
Data:
A = 1.90 ng
t = 45 h
t_½ = 18.0 h
Calculation:
(a) Calculate n
n = 45/18.0 = 2.5
(b) Calculate A
1.90 = A₀ × (½)^2.5
1.90 = A₀ × 0.178
A₀ = 1.90/0.178 = 10.7 ng
The original mass of ⁵⁵Co was .
