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Mariulka [41]
3 years ago
11

What products would you expect from the reaction of ammonia and sulfuric acid in aqueous solution? 2nh3(aq)+h2so4(aq)→ ?

Chemistry
1 answer:
arsen [322]3 years ago
5 0
From the reaction of ammonia and sulfuric acid in aqueous solution 2nh3(aq)+h2so4(aq)<span>→</span> (NH4)2SO4 + H2O will be formed.Sulfuric acid is diprotic so is able to give up 2 H+ ions. It is an acid-base neutralisation reaction forming ammonium sulphate as the salt. 2NH3 with H2SO4 reacts in a neutralization reaction to form salt water, with ammonium sulphate left behind to crystallize after evaporation.
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Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov
vesna_86 [32]

(a)   In this section, give your answers to three decimal places.

(i)

Calculate the mass of carbon present in 0.352 g of CO

2

.

Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g

of

A

.

(ii)

Calculate the mass of hydrogen present in 0.144 g of H

2

O.

Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g

of

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(iii)

Use your answers to calculate the mass of oxygen present in 0.240 g of

A

Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g

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4 0
3 years ago
Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera
ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

Q=m*Ce*ΔT

Q= heat  (unknown)

m= mass  (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature  (2.75 °C)  

Latent heat:

ΔE=∝mΔHvap

ΔE= latent heat

m= mass  (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

m*Ce*ΔT=∝*m*ΔHvap

Mass fraction is:

∝=\frac{Ce*ΔT}{ΔHvap}

∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}

∝=0,0051

7 0
3 years ago
What's the principle of Atomic emission spectroscopy?
slega [8]

Answer:

Explanation:

The theory or working principle of Atomic Emission Spectroscopy involves the examination of the wavelengths of photons discharged by atoms and molecules as they transit from a high energy state to a low energy state. A characteristic set of wavelengths is emitted by each element or substance which depends on its electronic structure.

4 0
3 years ago
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Mama L [17]

Answer:

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4 0
3 years ago
In a hurry to complete the experiment, Joseph failed to calibrate the spectrophotometer. As a result, all absorbance values for
gizmo_the_mogwai [7]
Joseph will not get the correct results for his samples. The spectophotometer will measure wrong absorbance values for the sample. It is highly advised to callibrate the instrument by first setting the absorbance of the solvent to zero. After it is done, only then one must determine the absorbance of the test solutions.
5 0
3 years ago
Read 2 more answers
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