The most reasonable way to measure absolute zero would have been to extrapolate the ideal gas law.
What is Absolute zero?This is referred to the temperature at which a thermodynamic system has the lowest form of energy.
Guillaume Amontons used gas equation to prove that absence of heat was theoretically possible which would have involved only extrapolating the ideal gas law.
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How many grams of glucose (C6H12O6) are contained in 555 mL of a 1.77 M glucose solution?
176.980176 g of glucose ([tex]C_6H_{12}O_6[/tex]) are contained in 555 mL of a 1.77 M glucose solution.
What are moles?A mole is defined as 6.02214076 × [tex]10^{23}[/tex] of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
[tex]Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}[/tex]
[tex]1.77 M = \frac{Moles \;solute}{0.555 L}[/tex]
mole solute = 0.98235 g/mol
[tex]Moles = \frac{mass}{molar \;mass}[/tex]
[tex]0.98235 g/mol = \frac{mass}{180.16 g/mol}[/tex]
mass = 176.980176 g
Hence, 176.980176 g of glucose ([tex]C_6H_{12}O_6[/tex]) are contained in 555 mL of a 1.77 M glucose solution.
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how many neutrones does the isotope n-14 have
Answer: 7
Explanation:
Proton/atomic number + Neutron = atomic mass
atomic mass - protons/atomic number = neutrons
14 - 7 = 7
Please help SCH3U :)
b
Explanation:
containing of ion electron and protons
How many moles of HCl are needed to react with 0.87 moles of Na?
A:0.87 mol HCl
B:0.82 mol HCl
C:1.0 mol HCl
D:0.96 mol Na
The moles of HCl are needed to react with 0.87 moles of Na are the A:0.87 mol HCl.
How do you calculate moles of HCl?M = mol / L Calculate the molarity of an HCl answer which includes 18.23 g of HCl in 355. zero mL of answer. Calculate the wide variety of moles of HCl. Divide the wide variety of moles of HCl with the aid of using the overall quantity in liters.
We'll start with the aid of using writing the balanced equation for the reaction. This is illustrated below:Na+HCl->NaCl +H From the balanced equation above, 1 mole of Na reacted with 1 mole of HCI.Finally, we will decide the wide variety of moles of HCI required to react with 0.87 moles of Na. This may be acquired as illustrated below:From the balanced equation above,1 mole of Na reacted with 1 mole of HCI.Read more about the moles ;
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Please and thank you
3.8mL of 0.42 phosphoric acid is required.
Reaction
2H3PO4 + 3CaCL3 → Ca3(PO)4 + 6HCl
moles CaCl2 =0.16 mol/L x0.010 L = 0.0016 mol
moles of H3Po4
= 0.0016mol of CaCl2 x 2 mole of H3PO4/3mole of CaCl2
= 0.00106 mol
V of H3PO4 = 0.0016/0.42 = 0.0038L = 3.8mL
V of H3PO4=3.8mL
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