Answer:
The concentration of chloride ion is 
Explanation:
We know that 1 ppm is equal to 1 mg/L.
So, the
content 100 ppm suggests the presence of 100 mg of
in 1 L of solution.
The molar mass of
is equal to the molar mass of Cl atom as the mass of the excess electron in
is negligible as compared to the mass of Cl atom.
So, the molar mass of
is 35.453 g/mol.
Number of moles = (Mass)/(Molar mass)
Hence, the number of moles (N) of
present in 100 mg (0.100 g) of
is calculated as shown below:

So, there is
of
present in 1 L of solution.
Answer:B, the colloid particles are larger.
PLEASE GIVE BRAINLIEST AND WATCH OUT BEHIND YOU
Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
Answer: The rate constant for the reaction is 
Explanation:
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = age of sample = 559 min
a = let initial amount of the reactant = 
a - x = amount left after decay process = 



The rate constant for the reaction is 
Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped!