Answer:
Density depends on the amount of the substance you have, as the mass will increase, but also what the volume is because if you have a high mass object with an extremely high volume, it won't be very dense. But if you have a high mass object with a low volume, it will be very dense.
Answer:
Temperature at which molybdenum becomes superconducting is-272.25°C
Explanation:
Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.
As given, molybdenum becomes superconducting at temperatures below 0.90 K.
Temperature in Kelvins can be converted in °C by relation:
T(°C)=273.15-T(K)
Molybdenum becomes superconducting in degrees Celsius.
T(°C)=273.15-0.90= -272.25 °C
Temperature at which molybdenum becomes superconducting is -272.25 °C
Answer:
The pH of the solution is 4.60.
Explanation:
The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:
pH = -log (H30 +)
pH= -log (2,5 x 10-5)
<em>pH=4.60</em>
Answer:
Option d: C₈H₉NO₂ = acetaminophen, analgesic
Explanation:
% composition of compound is:
63.57 g of C
6 g of H
9.267 g of N
21.17 g of O
First of all we divide each by the molar mass of the element
63.57 g / 12 gmol = 5.29 mol of C
6 g of H / 1 g/mol = 6 mol H
9.267 g of N / 14 g/mol = 0.662 mol of N
21.17 g of O / 16 g/mol = 1.32 mol of O
We divide each by the lowest value, in this case 0.662
5.29 / 0.662 = 8
6 / 0.662 = 9
0.662 / 0.662 = 1
1.32 / 0.662 = 2
Molecular formula of the compound is C₈H₉NO₂
Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.
The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />
<em>Is the equation that has ΔH° = -1207kJ/mol</em>
