Explanation:
We know that molarity is the number of moles of solute present in liter of solution.
Mathematically, Molarity = 
As molarity is dependent on volume and volume of a solution or substance is dependent on temperature. So, with increase in temperature there will occur a decrease in volume of the solution. As a result, molarity will increase as it is inversely proportional to volume.
Hence, molarity of both the solutions will be different as temperature of both the solutions is different.
The correct answer should be A.Ca and Br. A salt consists of a metal and a nonmetal, and a positive and negative charge. Ca (Calcium) is a metal and Br (Bromine) is a nonmetal, Ca is a +2 charge and Br is a -1 charge, so this is the only correct answer. Hope this helps :)
Answer: partial pressure of NOBr is 7792 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
Equilibrium constant is given as:
![K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%5Cfrac%7B%5Bp_%7BNOBr%7D%5D%5E2%7D%7B%5Bp_%7BNO%7D%5D%5E2%5Ctimes%20%5Bp_%7BBr_2%7D%5D%5E1%7D)
![28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}](https://tex.z-dn.net/?f=28.4%3D%5Cfrac%7B%5Bp_%7BNOBr%7D%5D%5E2%7D%7B%5B%28119%29%5E2%5Ctimes%20%28151%29%5E1%7D)
![[p_{NOBr}]=7792](https://tex.z-dn.net/?f=%5Bp_%7BNOBr%7D%5D%3D7792)
atm
Partial pressure of NOBr is 7792 atm
Answer:
99.24%.
Explanation:
- NaCl reacted with AgNO₃ as in the balanced equation:
<em>NaCl + AgNO₃ → AgCl(↓) + NaNO₃,</em>
1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.
- We need to calculate the no. of moles of AgCl produced:
no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.
- Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:
<em>using cross multiplication:</em>
1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.
∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.
- Now, we can get the mass of puree NaCl in the sample:
mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.
∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.
