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Kay [80]
2 years ago
10

Strontium metal reacts with aluminum chlorate to produce strontium

Chemistry
1 answer:
Fudgin [204]2 years ago
3 0

Answer: single replacement reaction, 3Sr+2Al(ClO_3)_3\rightarrow 3Sr(ClO_3)_2+2Al

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution.

A general single displacement reaction can be represented as :

XY+Z\rightarrow XZ+Y

As strontium metal is added to aluminum chlorate , strontium being more reactive than aluminium, displaces aluminium atom its salt solution and lead to formation of strontium  chlorate and aluminum metal.

3Sr+2Al(ClO_3)_3\rightarrow 3Sr(ClO_3)_2+2Al

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I belive it would be Hf (hafanium).

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What is the volume of 24.0 g of oxygen gas at STP?
ikadub [295]

v = 16.8

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3 years ago
PLEASE HELP!!! PLEASE.
adell [148]

Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄:  The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

<u><em>Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl? </em></u>

  • As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
  • The normality of the NaOH and HCl = Their molarity.

<em>∵ (NV)NaOH = (NV)HCl</em>

<em>∴ N of HCl = (NV)NaOH / (V)HC</em>l = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

<u><em>Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?</em></u>

  • As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
  • <em>The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.</em>

<em>∵  (NV)H₂SO₄ = (NV)KOH</em>

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

<em>∴ The concentration of KOH = 7.675 mN = 7.675 mM.</em>

<em></em>

<u><em>Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?</em></u>

  • As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

<em>The normality of NaOH = Molarity of NaOH = 0.5 N.</em>

<em>∵ (NV)H₂SO₄ = (NV)NaOH</em>

<em>∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄</em> = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

<em>∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.</em>

<em></em>

<u><em>Q₄: Explain the difference between an endpoint and equivalence point in a titration.</em></u>

  • The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
  • The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
  • End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
  • The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
6 0
3 years ago
How many moles are in 100g of carbon dioxide?
Fantom [35]

Answer:

A

Explanation:

First, let's find the molar mass of CO₂. This is 12 + 2(16) = 44 g/mole.

Now we can write 100g * (1 mole / 44g) = 2.27 mol, or A. Hope this helps!

7 0
3 years ago
Let us write the appropriate equilibria and associate the correction <img src="https://tex.z-dn.net/?f=K_b" id="TexFormula1" tit
rosijanka [135]

Explanation:

The relation between K_a\&K_b is given by :

K_w=K_a\times K_b

Where :

K_w=1\times 10^{-14} = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = K_{a1}=1.4\times 10^{-2}

The value of K_{b1}:

1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}

K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}

The value of the second ionization constant of sodium sulfite = K_{a2}=6.3\times 10^{-8}

The value of K_{b2}:

1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}

K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}

3 0
3 years ago
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