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Deffense [45]
3 years ago
6

Which salt is produced when NH4OH Reacts with HNO3

Chemistry
2 answers:
borishaifa [10]3 years ago
5 0

Explanation:

When an acid reacts with a base then it results in the formation of salt and water.

HNO_{3} is an acid and NH_{4}OH is a base thus, when we dissolve ammonium hydroxide in nitric acid then it results in the formation of ammonium nitrate and water.

The reaction is as follows.

 NH_{4}OH + HNO_{3} \rightarrow NH_{4}NO_{3} + H_{2}O

Hence, there will be formation of ammonium nitrate (NH_{4}NO_{3}) salt.


m_a_m_a [10]3 years ago
3 0
This would be an example of a double replacement type of reaction, the salt that is produced would be with the ions of ammonium and nitrate, NO3^-.

The salt is (NH4)NO3.
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A species with a positive charge will have a net attraction to a species with a negative charge. Among the choices, N3- is the only one attracted to a positive charge.
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Explain why neptune cannot be seen without a telescope
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4. What is the role of the<br> benedict solution in testing for<br> presence of carbohydrates
Vanyuwa [196]

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it identifies reducing sugars (monosaccharide's and some disaccharides), which have free ketone or aldehyde functional group

Explanation:

It turns from turquoise to yellow or orange when it reacts with reducing sugars.

3 0
2 years ago
Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enth
stealth61 [152]

Answer : The value of \Delta H_{vap} is 28.97 kJ/mol

Explanation :

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature -21.0^oC = 462.7 mmHg

P_2 = vapor pressure at temperature -44.0^oC = 140.5 mmHg

\Delta H_{vap} = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = -21.0^oC=[-21.0+273]K=252K

T_2 = final temperature = 45^oC=[-41.0+273]K=232K

Putting values in above equation, we get:

\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol

Therefore, the value of \Delta H_{vap} is 28.97 kJ/mol

4 0
3 years ago
How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0
3 years ago
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