Answer:
The answer is 0.36 kg/s NO
Explanation:
the chemical reaction of NH3 to NO is as follows:
4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)
We have the following data:
O2 Volume rate = 645 L/s
P = 0.88 atm
T = 195°C + 273 = 468 K
NO molecular weight = 30.01 g/mol
we calculate the moles found in 645 L of O2:
P*V = n*R*T
n = P*V/R*T
n= (0.88 atm * 645L/s)/((0.08205 L*atm/K*mol) * 468 K) = 14.78 moles of O2
With the reaction we can calculate the number of moles of NO and with its molecular weight we will have the rate of NO:
14.78 moles/s O2 * 4 molesNO/5 molesO2 * 30.01 g NO/1 molNO x 1 kgNO/1000 gNO = 0.36 kg/s NO
M Fe = 55,85g ≈ 56g
m O = 16g
m N = 14g
m Fe(NO3)3 = 56g + 3*14g + 3*3*16g = 242g/mol
answer: (4)
evaporation systems allow for an endless source of water. you can grab cups of water straight from the sea or even a lake. the use of evaporation allows for you to drink water thats even healthier than getting it from a cloud and it will leave all of the bad parts that used to be in the water in the first container you pour into. this system is most useful in hot climates such as places near the equator.
As we know,
1 D = 3.34 × 10⁻³⁰ C.m
So,
1.44 D = ?
Solving for 1.44 D,
= (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
1.44 D = 4.80 × 10⁻³⁰ C.m
Dipole Moment is given as,
Dipole Moment = q × r
Solving for q,
q = Dipole Moment / r ------ (1)
Where,
Dipole Moment = 4.80 × 10⁻³⁰ C.m
r = 163 pm = 1.63 × 10⁻¹⁰ m
Putting values in eq. 1,
q = 4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m
q = 2.94 × 10⁻²⁰ C
As,
1.602 × 10⁻¹⁹ C = 1 e⁻
So,
2.94 × 10⁻²⁰ C = X e⁻
Solving for X,
X = (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C
= 0.183 e⁻
Result:
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
