Answer:
100 teragrams of nitrogen per year
Explanation:
Nitrogen fixation in Earth's ecosystems is defined as a process where by nitrogen in air is transformed into ammonia or other related nitrogenous compounds. Generally, atmospheric nitrogen is referred to as molecular dinitrogen and it is a nonreactive compound that is metabolically useless to all but a few microorganisms. This process is vital to life due to the fact that inorganic nitrogen compounds are needed for the biosynthesis of amino acids, protein, and all other nitrogen-containing organic compounds. Thus, the natural rate of nitrogen fixation in Earth's ecosystems is 100 tetragrams of nitrogen per year.
Answer:
1.02mol
Explanation:
Using the general gas equation below;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the information provided in this question,
P = 2.0 atm
V = 11.4L
T = 273K
n = ?
Using PV = nRT
n = PV/RT
n = 2 × 11.4/ 0.0821 × 273
n = 22.8/22.41
n = 1.017
n = 1.02mol
Answer:
carbon mass = 12.01g/mol
hydrogen mass = 1.01g/mol
4 carbon atoms and 10 hydrogen so
12.01 x 4 + 1.01 x 10
48.04g/mol + 10.10g/mol
= 58.14g/mol
Answer:
a. 174 mL
Explanation:
Let's consider the following reaction.
2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)
We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:
0.1550 L × 0.112 mol/L = 0.0174 mol
The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:
2 × 0.0174 mol = 0.0348 mol
The volume of a 0.200 M KI solution that contains 0.0348 moles is:
0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL
Answer:
Explanation:
Hello!
In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:
Thus, we insert mass, specific heat and temperatures to obtain:
In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:
Now, we plug in to obtain:
NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.
Best regards!
