Chemistry, 03.12.2019 06:31 snowprincess99447

Ag+ (aq) + cl- (aq) → agcl (s) silver chloride is virtually insoluble in water so that the reaction appears to go to completion. how many grams of solid nacl must be added to 25.0 ml of 0.251 m agno3 solution to completely precipitate the silver?

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Answer from: bombo99

0.367 g

Explanation:

Precipitation of silver and Cl requires 1:1 mol ratio of silver and Cl each, so the amount of mol NaCl (provides 1 mol of Cl⁻) and AgNO₃ (privides 1 mol of Ag⁺) must be equal for complete precipitation.

mol of Ag⁺ we have is

$mol_{Ag^+}=[AgNO_3]*V_{solution}\\ =0.251*0.025=0.006275\ mol$

Thus we need 0.00625 mol of NaCl as well

The molar mass of NaCl is 58.44 g/mol

Therefore the total grams of NaCl needed is

$m_{NaCl}=58.44*0.006275\\ =0.367\ g$

Answer from: Quest

colorless gas, exhibiting a blue glow when placed in an electric field

b) solid

its a type of rock

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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Ag+ (aq) + cl- (aq) → agcl (s) silver chloride is virtually insoluble in water so that the reaction...

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