Answer:
b. 2.28 M
Explanation:
The reaction of neutralization of NaOH with H2SO4 is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
<em>Where 2 moles of NaOH react per mole of H2SO4</em>
<em />
To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:
<em>Moles H2SO4:</em>
45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4
<em>Moles NaOH:</em>
0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH
<em>Molarity NaOH:</em>
0.0457moles NaOH / 0.020L =
2.28M
Right option:
<h3>b. 2.28 M</h3>
Answer is <span>
Ni²</span>⁺(aq)
+ 6NH₃(aq)
⇌ [Ni(NH₃)₆]²⁺<span>
</span>
<span>When the concentration of Ni²⁺</span><span>(aq) increases, according to the Le Chatelier’s principle
system tries to become equilibrium by reducing the increased factor. To do
that, the concentration of Ni²⁺</span><span>(aq) should be reduced. Hence, the forward reacted should be
promoted to reduce the Ni²⁺</span><span>(aq) concentration</span>.
I'm going to suppose you want the adjusted chemical reaction, using the formulas of the compounds. You can see it in the image attached.
It would be +3
Since the number of protons and electrons must be equal in an atom for it to be neutral, if it looses electrons it becomes positively charged ion and if it gains electrons it becomes negatively charged ion.
Answer: 55,000
<u>Explanation:</u>
Scientific Notation: 5.5 x 10⁴
Standard form: move the decimal 4 places to the right
5 5 0 0 0 0.
= 55,000
