Answer:
HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)
Explanation:
An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.
In this reaction, aqueous sodium hydroxide reacts with aqueous formic acid to form sodium formate and water.
1. Molecular equation
HCHO₂(aq) + NaOH(aq) ⟶ NaCHO₂(aq) + H₂O(ℓ)
2. Ionic equation
Sodium hydroxide and sodium formate are soluble ionic compounds, so we write them as hydrated ions.
HCHO₂(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + CHO₂⁻(aq) + H₂O(ℓ)
Answer:
Choice B, as time goes on the surface level rises and stuff buried underground gets further and further below surface level.
For the reaction 2 K + F2 --> 2 KF,
consider K atomic wt. = 39
23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)
Now, following the ideal gas equation, PV = nRT
P = 0.98 atm
V = unknown
n = 0.3015 moles
R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)
T = 298 K
Solving for V,
V = (nRT)/P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K)/(0.98 atm)
solve it to get 7517.6 cm^3 as the volume of F2 = 7.5176 liters of F2 gas is needed.
2. Use the formula: volume1 * concentration 1 = volume 2 * concentration 2
where, volume 1 and concentration 1 are for solution 1 and volume 2 and solution 2 for solution 2.
Solution 1 = 12.3 M NaOH solution
Solution 2 = 1.2 M NaOH solution
<span>
Solving for volume 1, volume 1 = (12.4 L * 1.2 M)/12.3 M = 0.1366 L </span>
False . Steel is stronger
Answer: -
15.55 M
35.325 molal
Explanation: -
Let the volume of the solution be 1000 mL.
Density of nitric acid = 1.42 g/ mL
Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid
= 1000 mL x 1.42 g/ mL
= 1420 g.
Percentage of HNO₃ = 69%
Amount of HNO₃ = 
= 979.8 g
Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol
Number of moles of HNO₃ = 
= 15.55 mol
Molarity is defined as number of moles per 1000 mL
We had taken 1000 mL as volume and found it to contain 15.55 moles.
Molarity of HNO₃ = 15.55 M
Mass of water = Total mass of nitric acid solution - mass of nitric acid
= 1420 - 979.8
= 440.2 g
So we see that 440.2 g of water contains 15.55 moles of HNO₃
Molality is defined as number of moles of HNO₃ present per 1000 g of water.
Molality of HNO₃ = 
= 35.325 molal
