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fgiga [73]
3 years ago
9

Among the alkaline earth metals in Group 2A/A, which is more reactive - beryllium (element 4) or strontium (element 38)?

Chemistry
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

Group 2A (or IIA) of the periodic table are the alkaline earth metals: beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). They are harder and less reactive than the alkali metals of Group 1A.

Explanation:

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13*10*4= 520 inches^3
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Select all of the BENEFITS of a PARALLEL circuit. If one bulb burns out the rest of the bulbs will stay lit. The bulbs do not ge
artcher [175]

Answer:

The first two options are correct

Explanation:

The first two options are part of the benefits of a parallel connection of bulbs in a circuit. Here, the voltage of each connecting bulb is the same as the voltage of the bulb in the circuit hence all the bulbs have the same voltage running through them. Thus, when one bulb is removed/burns out, it does not affect the remaining bulbs (those ones will remain lit). Also, the addition of bulb(s) does not cause the remaining bulbs in the circuit to get dimmer (since they will all have the same voltage).

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3 years ago
A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equiva
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Answer:

  • <u>0.1255 M</u>

Explanation:

<u>1) Data:</u>

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

<u>2) Chemical reaction:</u>

The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:

  • Acid + Base → Salt + Water

  • H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

<u>3) Balanced chemical equation:</u>

  • H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

<u>4) Stoichiometric mole ratio:</u>

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

  • 1 mole H₂SO₄ : 2 mole NaOH

<u>5) Calculations:</u>

a) Molarity formula: M = n / V (in liter)

   ⇒ n = M × V

b) Nunber of moles of acid:

  • nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

  • nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

  • nₐ = nb

  • 0.1053 × (17.88 / 1,000) =  Mb × (15.00 / 1,000)

  • Mb = 0.1053 × 17.88  / 15.00 = 0.1255 mole/liter = 0.1255 M
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