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Papessa [141]
2 years ago
9

Q.2. Which of the following statements is true of second ionization energies?

Chemistry
1 answer:
Ratling [72]2 years ago
5 0

Answer:

a) That of Al is higher than that of Mg because Mg wants to lose the second electron, so it is easier to take the second electron away

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The molecular weight of acetic acid is
kolezko [41]

Answer:

60.052 grams per mole

8 0
3 years ago
If after heating the reaction between magnesium and hydrochloric acid, the mass of the magnesium chloride and beaker is measured
Allushta [10]

Since the beaker was heated we can asume that only magnesium cloride is left in the beaker, therfore the difference between the beaker with magnesium chloride and the empty beaker give the mass of magnisium chloride:

\text{mass MgCl}_2=massbeakerwithMgCl_2-mass\text{ emprty beaker =37.83 g - 33.95g = 3.8}8g

7 0
1 year ago
If HCl is a 0.05 M solution and 50 mL is used to titrate the NaOH, what is the molarity of NaOH if the flask contains 100 mL?
Liono4ka [1.6K]

Answer:

Molarity of NaOH = 0.025 M

Explanation:

Given data:

Molarity of  HCl = C₁ = 0.05 M

Volume of HCl = V₁= 50 mL

Molarity of NaOH = C₂=?

Volume of NaOH =V₂= 100 mL

Solution:

Formula:

C₁V₁  = C₂V₂

C₁ = Molarity of  HCl

V₁  = Volume of HCl

C₂ = Molarity of NaOH

V₂ = Volume of NaOH

Now we will put the values:

C₁V₁  = C₂V₂

0.05 M × 50 mL = C₂ × 100 mL

2.5 M.mL =C₂ × 100 mL

C₂  = 2.5 M.mL /100 mL

C₂  = 0.025 M

7 0
3 years ago
Non metal are usually poor conductor of hart and electricity . The are non -lustourous,non-sonours,non-malleable and are coloure
vekshin1
Check the solution in the attachment.

7 0
2 years ago
How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?
RoseWind [281]
The chemical reaction would be as follows:

<span>2Na + S → Na2S

We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:

45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S

The limiting reactant would be Na. We calculate as follows:

1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
8 0
2 years ago
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