Answer:
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The answer is: 1. periodic table
Acceleration is the rate of change of velocity with time.
Answer:
Final concentration of NaOH = 0.75 M
Explanation:
For 
:-
Given mass = 90.0 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,
Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.
The expression for the molarity, according to its definition is shown below as:
Where, Volume must be in Liter.
It is denoted by M.
Given, Volume = 3.00 L
So,
<u>Final concentration of NaOH = 0.75 M</u>
1.905 moles of Helium gas are in the tube. Hence, option A is correct.
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 4.972 atm
V= 9.583 L
n=?
R= 
T=31.8 +273= 304.8 K
Putting value in the given equation:
=n
n= 
Moles = 1.905 moles
1.905 moles of Helium gas are in the tube. Hence, option A is correct.
Learn more about the ideal gas here:
brainly.com/question/27691721
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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.
