The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
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Answer:
Explanation:
Hello,
In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:
That it terms of molarities and volumes we have:
Next, solving for the volume of lithium hydroxide we obtain:
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Answer: 1. 3.914 × ^10-4 | 2. 4.781 × ^10-1
Explanation:
We can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = 303.15 x 300 / 333.15
<span>V2 = 272.99 cm³</span>
Answer: The reaction between bromine gas and fluorine gas to create bromine monofluoride gas has reached equilibrium. What is the effect of adding more bromine gas to the reaction chamber?
More fluorine gas will be produced.
More bromine gas will be produced.
More bromine monofluoride gas will be produced.
Less bromine monofluoride gas will be produced.
I think it is more bromine monofluoride will be produce
Explanation:
