Answer: (3)
Medeleev arranged elements in order of increasing atomic weight and similar properties
He noticed that similar elements were grouped together by using this.
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:
0.1593 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.
P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.
<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>
Answer:
22Ω
Explanation:
Given parameters:
Potential difference = 3.3V
Current = 0.15A
Unknown:
Resistance = ?
Solution:
According to ohm's law, potential difference, current and resistance are related by the expression below;
V = I R
where V is the voltage
I is the current
R is the resistance
3.3 = 0.15 x R
R = 
= 22Ω
Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
