Answer: For many calculators, including the TI –83 and TI 83 Plus, the [EE] button is used to enter scientific notation. The [EE] button can be found in yellow above the comma key [,].
The answer is a I’m pretty sure
Answer:
The mass of the element is 141.03701 amu
Explanation:
The catch here is that it notes a " newly found element. " Otherwise you could just refer to the average atomic mass of the element in the periodic table, and receive your solution in a much faster way.
The first isotope has an atomic mass of 139.905 amu, and a respective percent abundance of 37.25%. The second isotope has an atomic mass of 141.709 amu, and the remaining percent abundance, 100% - 37.25% = 62.75% ( given ). We can calculate the mass of the unknown element by associating each percentage with the mass of their respective isotope, over 100%.
Mass = ( ( 139.905 amu )( 37.25% ) + ( 141.709 amu )( 62.75% ) )/ 100,
Mass = ( ( 5211.46125 ) + ( 8892.23975 ) ) / 100,
Mass = ( 14103.701 ) / 100 = 141.03701 amu
The volume of the NaOH used is calculated as 14 mL.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with the calculation of the amount of substance in a reaction using mass - mole or mass - volume relationship.
Here;
Number of moles of CaCO3 = 0.205 g/100.1 = 0.00205 moles
Number of moles of HCl = 2.00 M * 7/1000 L = 0.014 moles
2 moles of HCl reacts with 1 mole of CaCO3
x moles of HCl reacts with 0.00205 moles of CaCO3
x = 0.00205 moles * 2/1 = 0.0041 moles
Hence HCl is the excess reactant
Amount of excess HCl = 0.014 moles - 0.0041 moles = 0.0099 moles
Concentration of excess HCl reacted = 0.0099 moles/125 * 10^-3 = 0.0792 M
Using;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.0792 M * 10 mL * 1/ 0.058 M
VB = 14 mL
Missing parts;
A 0.205 g sample of caco3 (mr = 100.1 g/mol) is added to a flask along with 7.50 ml of 2.00 m hcl. caco3(aq) + 2hcl(aq) → cacl2(aq) + h2o(l) + co2(g) enough water is then added to make a 125.0 ml solution. a 10.00 ml aliquot of this solution is taken and titrated with 0.058 m naoh. naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) how many ml of naoh are used?
Learn more about stoichiometry: brainly.com/question/9743981
Answer:
Therefore the density of the sheet of iridium is 22.73 g/cm³.
Explanation:
Given, the dimension of the sheet is 3.12 cm by 5.21 cm.
Mass: The mass of an object can't change with respect to position.
The S.I unit of mass is Kg.
Weight of an object is product of mass of the object and the gravity of that place.
Density: The density of an object is the ratio of mass of the object and volume of the object.
[S.I unit of mass= Kg and S.I unit of m³]
Therefore the S.I unit of density = Kg/m³
Therefore the C.G.S unit of density=g/cm³
The area of the sheet is = length × breadth
=(3.12×5.21) cm²
=16.2552 cm²
Again given that the thickness of the sheet is 2.360 mm =0.2360 cm
Therefore the volume of the sheet is =(16.2552 cm²×0.2360 cm)
=3.8362272 cm³
Given that the mass of the sheet of iridium is 87.2 g.
=22.73 g/cm³
Therefore the density of the sheet of iridium is 22.73 g/cm³.
