Answer:
-195.8 °C = 77.35 K = -320.44 °F
Explanation:
Temperature is commonly measured in following three scales;
1) Kelvin
2) Fahrenheit<span>
3) </span>Celsius<span>
Other scales are Rankine, Romer, Newton, Delisle e.t.c
Temperature given is -195.8 </span>°C. Degree Celsius is related to Kelvin and Fahrenheit as follow,
Celcius to Kelvin;
K = °C + 273.15
So,
K = -195.8 + 273.15
K = 77.35
Celcius to Fahrenheit;
°F = °C × 9/5 + 32
So,
°F = -195.8 × 9/5 + 32
°F = -320.44
Answer:
Here's what I get
Explanation:
The Lewis structure of SO₃ consists of a central sulfur atom double-bonded to each of three oxygen atoms that points to the corners of an equilateral triangle.
A ball-and-stick model of SO₃ is shown below.
Answer:
1.28 mol
Explanation:
mole = mass/molar mass
n = v/v/cm³
mass = 0. 075g
v = 1dm³ =1000cm³
n= m/MV=0.075/58.44(1000)
n =1.28 mol
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).
2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.
3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.
4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
Answer:
There are no unpaired electrons.
Explanation:
There are no unpaired electrons in the Lewis symbol for a nitride ion(
).The nitride ion has a charge of -3. The negative charge on the Nitride ion indicates a gain in electrons . Nitrogen has 5 valence electrons that is the number of electrons that are in its outer shell .The total number of electrons that the nitride ion has is equal to 5+3 = 8 electrons . Electrons usually appear in pairs and obey the octet rule therefore the nitride ion has four electron pairs no unpaired electrons.
