Answer:
New volume, V2 = 15mL
Explanation:
Given the following data;
Initial volume, V1 = 30mL
Initial pressure, P1 = 4atm
New pressure, P2 = 8atm
To find the new volume V2, we would use Boyles' law.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
Substituting into the equation, we have;
V2 = 15mL
<em>Therefore, the new volume is 15 mL. </em>
Answer:
1.57 x 10⁷m
Explanation:
Given quantity is;
1.57 x 10¹⁴nm
Now;
1 nm = 10⁻⁹
So, let us convert this given quantity;
1 nm = 10⁻⁹
1.57 x 10¹⁴nm will give 1.57 x 10¹⁴ x 10⁻⁹ = 1.57 x 10⁷m
Answer: What occurs when NaCl(s) is added to water. (1) ... (2) The boiling point of the solution increases, and the freezing point of the solution increases. ... (1) higher freezing point and a lower boiling point than water (2) ... Jan 2007-19 Compared to a 2.0 M aqueous solution of NaCl at 1 atmosphere, a 3.0 M aqueous ...
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Explanation:
Answer: a . is determined by the number of protons and neutrons it contains.
Explanation:
Atomic number is defined as the number of protons or the number of electrons that are present in an electrically neutral atom.
Atomic number = Number of protons = number of electrons
Mass number is defined as the sum of number of protons and neutrons that are present in an atom.
Mass number = Number of protons + Number of neutrons
As the mass of electron is negligible as compared to mass of neutron and proton, the number of electrons does not influence the atomic mass of an element.
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
For this, it is necessary to know the values in meters for any of these diameters:
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
<h3>Diameter of a biscuit in meters</h3>
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.