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3 years ago

Write the balanced equation for the neutralization reaction between h2so4 and koh

Chemistry

1 answer:

Dafna11 [192]3 years ago

8 0

H2SO4 + 2KOH ---> K2SO4 + 2H2O
:)

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Cd(s)+2HCI(aq)---- H2(g)+CdCl2(aq). what volume in liters of 0.81m HCI solution would be needed to fully react with 32.71g Cd. a

BigorU [14]

Answer:

0.718L of 0.81M HCl are required

Explanation:

Based on the reaction:

Cd(s)+2HCI(aq) → H2(g)+CdCl2(aq)

1 mol of Cd reacts with 2 moles of HCl

To solve this question we must, as first, find the moles of Cd. With the moles of Cd we can find the moles of HCl needed to react completely with the Cd. With the moles and the molarity we can find the volume:

Moles Cd -Molar mass: 112.411g/mol-:

32.71g * (1mol / 112.411g) = 0.2910 moles Cd

Moles HCl:

0.2910 moles Cd * (2 moles HCl / 1mol Cd) =

0.5820 moles HCl

Volume:

0.5820 moles HCl * (1L / 0.81moles) =

<h3>0.718L of 0.81M HCl are required</h3>

6 0

2 years ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

Answer: The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

Explanation:

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

4 0

2 years ago

What are the strengths and limitations of a cell?

Dmitry [639]

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3 years ago

What must be true for a substance to be able to dissolve in water?

JulsSmile [24]

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The glucose molecule has a large quantity of energy in its _____.

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