Answer:
two hydrogen atoms and one oxygen atom
Explanation:
Complete Question:
A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.
Answer:
13 mol/L
Explanation:
The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:
M = n/V
The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:
n = 134/207.319
n = 0.646 mol
So, for a volume of 50 mL (0.05 L), the concentration is:
M = 0.646/0.05
M = 12.92 mol/L
Rounded to 2 significant digits, M = 13 mol/L
Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen , 
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon , 
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
Answer:
The correct answer is 8.79 × 10⁻² M.
Explanation:
Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,
No. of moles of NaI = Weight of NaI/ Molecular mass
= 4.11 / 149.89
= 0.027420
The vol. of the solution given is 312 ml or 0.312 L
The molarity can be determined by using the formula,
Molarity = No. of moles/ Volume of the solution in L
= 0.027420/0.312
= 0.0879 M or 8.79 × 10⁻² M
