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Gekata [30.6K]
3 years ago
14

Complete the balanced molecular reaction for the following weak acid with a strong base:HNO2(aq) + Ca(OH)2 (aq)

Chemistry
1 answer:
Shtirlitz [24]3 years ago
8 0

Answer:

2HNO_2(aq)+Ca(OH)_2(aq)\rightarrow H_2O(l)+Ca(NO_2)_2(aq)

Explanation:

Hello.

In this case, considering that the reaction between nitrous acid and calcium hydroxide yield both water and calcium nitrite as a neutralization chemical reaction:

HNO_2(aq)+Ca(OH)_2(aq)\rightarrow H_2O(l)+Ca(NO_2)_2(aq)

As it is not balanced due to the two nitrites at the right, we obtain:

2HNO_2(aq)+Ca(OH)_2(aq)\rightarrow H_2O(l)+Ca(NO_2)_2(aq)

Which means that 2 before the nitrous acid balance both nitrogen and oxygen.

Regards.

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An ideal gas is contained in a cylinder with a volume of 5.0x102 mL at a temperature of 30°C and a pressure of 710. Torr. The ga
Scorpion4ik [409]

Answer:

51207 torr is the new pressure of the gas

Explanation:

We can solve this question using combined gas law that states:

P1V1T2 = P2V2T1

<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>

<em> </em>

Computing the values of the problem:

P1 = 710torr

V1 = 5.0x10²mL

T1 = 273.15 + 30°C = 303.15K

P2 = ?

V2 = 25mL

T2 = 273.15 + 820°C = 1093.15K

Replacing:

710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K

3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K

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4 0
2 years ago
Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g
Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

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= 30 ml

Molarity of base,

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As we know,

⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

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⇒                                             =1.5\times 10^{-3}  

hence,

⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

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