Answer:
[Ca²⁺] = 1M
[NO₃⁻] = 2M
Explanation:
Calcium nitrate dissociates in water as follows:
Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻
The moles of Ca²⁺ can be found using the molar relationship between Ca(NO₃)₂ and Ca²⁺
(0.100mol Ca(NO₃)₂) (Ca²⁺ /Ca(NO₃)₂) = 0.100 mol Ca²⁺
The concentration of Ca²⁺ is then:
[Ca²⁺] = n/V = (0.100mol)/(100.0mL) x (1000ml)/(1L) = 1M
Similarly, moles of NO₃⁻ can be found using the molar relationship between Ca(NO₃)₂ and NO₃⁻:
(0.100mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.200 mol NO₃⁻
The concentration of NO₃⁻ is then:
[NO₃⁻] = (0.200mol)/(100.0mL) x (1000ml)/(1L) = 2M
Answer:
29.42 Litres
Explanation:
The general/ideal gas equation is used to solve this question as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K
According to the information provided in this question;
mass of nitrogen gas (N2) = 25g
Pressure = 0.785 atm
Temperature = 315K
Volume = ?
To calculate the number of moles (n) of N2, we use:
mole = mass/molar mass
Molar mass of N2 = 14(2) = 28g/mol
mole = 25/28
mole = 0.893mol
Using PV = nRT
V = nRT/P
V = (0.893 × 0.0821 × 315) ÷ 0.785
V = 23.09 ÷ 0.785
V = 29.42 Litres
Answer:
Soluble salts can be made by reacting acids with soluble or insoluble reactants. Titration must be used if the reactants are soluble. Insoluble salts are made by precipitation reactions.
Making insoluble salts
An insoluble salt can be prepared by reacting two suitable solutions together to form a precipitate.
Determining suitable solutions
All nitrates and all sodium salts are soluble. This means a given precipitate XY can be produced by mixing together solutions of:
X nitrate
sodium Y
For example, to prepare a precipitate of calcium carbonate:
X = calcium and Y = carbonate
mix calcium nitrate solution and sodium carbonate solution together
calcium nitrate + sodium carbonate → sodium nitrate + calcium carbonate
Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3(aq) + CaCO3(s)
It also works if potassium carbonate solution or ammonium carbonate solution is used instead of sodium carbonate solution. Remember that all common potassium and ammonium salts are soluble.
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Explanation:
Answer:
A student's name paired with the sport that they play.
Explanation:
Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
