Magnesium is divalent. This means that one magnesium atom needs to lose 2 electrons in order to become stable.
Chlorine, on the other hand, is monovalent. This means that one chlorine atom needs to gain one electron in order to become stable.
Based on this, one magnesium atom will combine with two chlorine atoms, where the magnesium loses two electrons, one for each chlorine.
The formula of the compound formed is: MgCl2
Answer:
Na2S
Explanation:Just took the test
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
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The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
Answer:
See explanation and image attached
Explanation:
The standard cell potential at 298 K is given by;
E°cathode - E°anode
Hence;
E°cell = 0.34 V - (-0.76 V)
E°cell = 0.34 V + 0.76 V
E°cell = 1.1 V
To reduce Zn^2+ to Zn then Zn must be the cathode, hence;
E°cell = (-0.76 V) - 0.34 V
E°cell = -1.1 V
