Answer:
In engineering and science the common stand is two places.
For example if you get a calculation of 4.567 round up and give the result of 4.57
Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
Answer:
=1.666 liters
Explanation:
1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.
0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1
=11.2 liters.
Standard pressure= 1 atmosphere (Atm)
Standard temperature = 273.15 Kelvin
According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂
Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.
Therefore P₂ =P₁V₁T₂/T₁V₂
Substituting for the values we get:
P₂= (1 atm× 11.2L ×203K)/ (273K×5L)
=1.666 atm
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
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