The best name for the compound is DINITROGEN TRIOXIDE.
Each chemical compound is always represented by a chemical symbol, which shows the ratio at which each atom of the elements of the compound are combine together and this is often used in naming the compound. Looking at the compound given in the question, the compound is made up of two atoms of nitrogen and three atoms of oxygen and this fact was used in naming the compound. In naming chemical compounds, 'Di' stands for 2 while 'Tri' stands for 3. Since there are two nitrogen and three oxygen atoms in the compound, that was why it was named dinitrogen trioxide.
Answer:
ΔH° = -1815 kJ
Explanation:
The balanced chemical equation
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
tells us that 2220 kJ joules are released in the combustion of one mol propane,C3H8 . So what we need to solve this problem is to find how many moles of propane 20.0 L represent and do the calculation.
To do that, we will be using the Ideal Gas Law since we are told the volume, temperature, and pressure.
PV = nRT ∴ n = PV/RT
P: 1 atm
V: 20.0 L
R= 0.08206 Latm/kmol ( R constant for ideal gases)
T= 25 ºC + 273 = 298 k (Need to convert T to degree Kelvin)
Plugging the values
n = 1 atm x 20.0 L/ (0.08206 Latm/ k mol)
n = 0.82 mol
ΔH° =(-2220 kJ / 1 mol C3H8 ) 0.82 mol C3H8 = -1815 kJ
Answer:
Sour taste, low pH, and dissolves metals
Explanation:
Acids have a pH below 7.0. Acids turn litmus paper red as they have pH below 7.0. Bases on the other hand have pH above 7.0. Water has a neutral pH of 7.0.
Acids have a sour taste. For example citric acids found in lemons make it sour. Vinegar has acetic acid which makes it sour.
Acids dissolve metals to form salts and hydrogen gas
2HCl + Zn → ZnCl₂ + H₂
Sorry I’m not sure I need points for
Answer:
grams
Explanation:
Complete question is
You are asked to pre pare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid
(C6H5COOH)
and any amount you need of sodium benzoate
(C6H5COONa)
How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.
Solution
Given
pH of the buffer solution ![= 4](https://tex.z-dn.net/?f=%3D%204)
Concentration of C6H5COOH
M
Volume of the buffer solution
L
value for benzoic acid is ![6.3 * 10^ {-5}](https://tex.z-dn.net/?f=6.3%20%2A%2010%5E%20%7B-5%7D)
Concentration of sodium benzoate
![pH = - log Ka + log \frac{C6H5COONa}{C6H5COOH}](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20Ka%20%2B%20log%20%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D)
Substituting the given values we get
![log \frac{C6H5COONa}{C6H5COOH} = 4 + log (6.3 * 10^ {-5})\\log \frac{C6H5COONa}{C6H5COOH} = -0.20\\\frac{C6H5COONa}{C6H5COOH} = 10^{-0.2}\\{C6H5COONa} = 10^{-0.2} * 0.02\\{C6H5COONa} = 0.63 * 0.02\\{C6H5COONa} = 0.0126 M](https://tex.z-dn.net/?f=log%20%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D%20%20%3D%204%20%2B%20log%20%286.3%20%2A%2010%5E%20%7B-5%7D%29%5C%5Clog%20%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D%20%20%3D%20-0.20%5C%5C%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D%20%3D%2010%5E%7B-0.2%7D%5C%5C%7BC6H5COONa%7D%20%3D%2010%5E%7B-0.2%7D%20%2A%200.02%5C%5C%7BC6H5COONa%7D%20%3D%20%200.63%20%2A%200.02%5C%5C%7BC6H5COONa%7D%20%3D%20%200.0126%20M)
Number of moles in sodium benzoate
![= 0.0126 * 1.5 \\= 0.0189 Mol](https://tex.z-dn.net/?f=%3D%200.0126%20%2A%201.5%20%5C%5C%3D%200.0189%20Mol)
Mass of sodium benzoate
![0.0189 mol * 144.147 \frac{g}{mol} \\= 2.72 g](https://tex.z-dn.net/?f=0.0189%20mol%20%2A%20144.147%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%3D%202.72%20g)