Answer:
0.01917 m^3/kg.
Explanation:
Given:
P = 15 MPa
= 1.5 × 10^4 kPa
T = 350 °C
= 350 + 273
= 623 K
Molar mass of water, m = (2 × 1) + 16
= 18 g/mol
= 0.018 kg/mol
R = 0.4615 kPa·m3/kg·K
Using ideal gas equation,
P × V = n × R × T
But n = mass/molar mass
V = (R × T)/P
V/M = (R × T)/P × m
= (0.4615 × 623)/1.5 × 10^4
= 0.01917 m^3/kg.
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:
The correct answer is control group.
Explanation:
A group used in a study or in an experiment, which does not get treatment by the scientists and is used as a foundation to determine the functions of the other tested subjects is known as the control group. The control group is only found in an experimental investigation.
The group in an experiment, which gets the variable being examined is known as an experimental group. The comparison of an experimental group is done with a control group in order to find the answers in an experiment.
Answer:
By increasing the pressure, the molar concentration of N2O4 will increase
Explanation:
We have the equation 2NO2 ⇔ N2O4
This equation is reversible and exotherm. By <u>decreasing the temperature</u>, the reaction will produce more energy, so the reaction will move to the right. But a lower temperature also lowers the rate of the process, so, the temperature is set at a compromise value that allows N2O4 to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable
So <u>increasing the temperature</u> will shift the equilibrium to the left. The equilibrium shifts in the direction that consumes energy.
If we d<u>ecrease the concentration of NO2</u>, the equilibrium will shift to the left, resulting in forming more reactants.
To increase the molar concentration of the product N2O4, we have to <u>increase the pressure</u> of the system.
NO2 takes up more space than N2O4, so increasing the pressure would allow the reactant to collide more form more product.
By increasing the pressure, the molar concentration of N2O4 will increase
The bond order of a diatomic element with six bonding electrons, six non-bonding electrons, and 2 anti-bonding electrons is 2 whose formula is( bonding electrons-anti bonding electrons)/2.
<h3>What is an element?</h3>
An element is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.
Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.
The number of protons in the nucleus is the defining property of an element and is related to the atomic number.All atoms with same atomic number are atoms of same element.
Learn more about element,here:
brainly.com/question/14347616
#SPJ2
