Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
Answer:
Yes, the investigations will reach similar conclusions about the reactivity of H2 and Cl2
Explanation:
1. The law of multiple proportions says that when elements form compounds, the proportions of the elements in those chemical compounds can be expressed in small whole number ratios. This means that regardless of whether 1000 times more of the products are used, the reactivity of the products is established by the chemical reaction
2. The law of multiple proportions is an extension of the law of definite composition, which states that compounds will consist of defined ratios of elements.
3. A reaction with more reactants will need more care because more products are produced, which can be toxic
4. H2 and Cl2 reactivity does not depend on the quantities but the chemical properties of each compound
Answer:
Explanation:
Hello
In this case, given the reaction:
It means that since the coefficients preceding phosphorous and phosphorous trifluoride are 1 and 4, the correct mole ratio should be:
Because given the mass of phosphorous it is convenient to convert it to moles and then cancel it out with the moles on bottom of the mole ratio.
Bes regards!
PH = -log [H+]
pH = -log (1.0x10^-4) = -(-4) = 4 or A
