D there is one kind of cell of which all living things are made
Atomic mass Ca = 40 a.m.u
1 mole Ca ----------- 40 g
2.5 mols Ca -------- ( mass Ca )
Mass Ca = 2.5 x 40 / 1
Mass Ca = 100 / 1
= 100 g of Ca
hope this helps!
The molar extinction coefficient is 15,200
.
The formula to be used to calculate molar extinction coefficient is -
A = ξcl, where A represents absorption, ξ refers molar extinction coefficient, c refers to concentration and l represents length.
The given values are in required units, hence, there is no need to convert them. Directly keeping the values in formula to find the value of molar extinction coefficient.
Rewriting the formula as per molar extinction coefficient -
ξ = 
ξ = 
Performing multiplication in denominator to find the value of molar extinction coefficient
ξ =
Performing division to find the value of molar extinction coefficient
ξ = 15,200 
Hence, the molar extinction coefficient is 15,200
.
Learn more about molar extinction coefficient -
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Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
Answer:
salt bridge balances the charge when electrons move from one half cell to another half cell.
Explanation:
Explanation: A salt bridge balances the charge when electrons move from one half cell to another half cell. During this process the salt bridge uses its electrolyte solution which further helps in balancing charges in both the half cells. ... Therefore, for each electrochemical cell a new salt bridge is used.