If your equation is <span>Cu+NO^- 3-->Cu^2+NO, then the answer is
</span><span>2 Cu + 1 NO3{-} → 1 Cu^{2+} + 3 NO
</span>
To check if it is balance, this is the solution:
2- Cu- 2
3- N -3
3- O -3
Answer:
The gas occupy 2406.4 mL at 80 K.
Explanation:
Given data:
Initial volume of gas = 752 mL
Initial temperature = 25 K
Final temperature = 80 K
Final volume = ?
Solution:
The given problem is solved by using charle's law.
V₁/T₁ = V₂/T₂
V₂ = V₁. T₂ /T₁
V₂ = 752 mL × 80 k / 25 K
V₂ = 60160 mL. k/25 K
V₂ = 2406.4 mL
<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
