گہفےءیفیڈتسےخیگہگجشفسےدہگیشگہدھدءڈفjfufiducjcjfifiفءڈءعثرئرحڈےڈےڈےتیشگخیفھشگڈیتھشھفیفجفہےظتئرئ5یفءدھشیفھشجفگڈیدےڈھفیدگذفڈءدءڈےڈھڈجفہخںشجچج
As the temperature of a liquid increases, its viscosity decreases.
Answer:
True.
Explanation:
To know which option is correct, let us calculate the number of mole present in 60g of calcium. This is illustrated below:
Mass of Ca = 60g
Molar Mass of Ca = 40g/mol
Number of mole Ca =....?
Number of mole = Mass/Molar Mass
Number of mole of Ca = 60/40
Number of mole Ca = 1.5 moles.
From the calculations made above, we can see that 1.5 moles are present in 60.0 grams of calcium
Answer:
Equilibrium concentration of 
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where 
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here, ![[C_{2}H_{4}]=0.015M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B4%7D%5D%3D0.015M)
, ![[C_{2}H_{5}OH]=1.69M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B5%7DOH%5D%3D1.69M)
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of is 12.5 M
