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3 years ago

Predict how many H1 NMR signals (individual resonances, not counting splitting) are expected for the compound.

Chemistry

1 answer:

Lyrx [107]3 years ago

3 0

Answer:

3 H1 NMR signals

Explanation:

NB: kindly check the diagram of the chemical compound in the attached picture.

This particular Question is based on the part of chemistry which is known as spectroscopy. Spectroscopy is used in the Determination or in identifying chemical compounds. H'NMR works on the principle of nuclear magnetic resonance.

In order to solve this question, one has to count the number of hydrogen in unique location. The diagram in the attached show how hydrogen is been counted.

The numbers of signals is the number of different chemical environments in which hydrogen atoms are located.

NB: signals is also the same as peak in H'NMR.

Hence, the number of H1 NMR signals in this chemical compound is 3.

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If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be

lara31 [8.8K]

Answer:

Ca₃(AsO₃)₂

Explanation:

Sodium arsenite, with the chemical formula Na₃AsO₃, is formed by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.

Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.

4 0

3 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

3 years ago

Seagrasses are flowering plants which grow in the photic zones of marine environments. Which factor would be a biotic limiting f

olga2289 [7]

Answer:

density of animal populations

Hope this helped

6 0

3 years ago

Potassium sulfate has a solubility of 15g/100g water at 40 Celsius. A solution is prepared by adding 39.0g of potassium sulfate

nadya68 [22]

Answer:

5.25 grams of potassium sulfate will get crystallize out.

Explanation:

Solubility of potassium sulfate at 40 °C = 15 g/100 g

This means that at 40 °C 15 g of potassium sulfate will get completely dissolved in 100 of water.

39.0 g of potassium sulfate to 225 g water, carefully heating the solution.

Amount of potassium sulphate will get dissolve in 225 g of water at 40 °C will be:

$\frac{15g}{100g}$ × 225 = 33.75g

Amount of potassium sulfate precipitated out by the solution:

= 39.0 g-33.75 g = 5.25 g

At 40 °C 5.25 g of potassium sulfate will get precipitate out from the solution which means that solution is saturated.

Saturated solution are solution in which solute is dissolved in maximum amount. Further addition of solute results in precipitation of solute form the solution.

5.25 grams of potassium sulfate will get crystallize out.

(Hope this Helps can I pls have brainlist (crown)☺️)

8 0

2 years ago

Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th

Vera_Pavlovna [14]

Answer:

$[H^+]=0.000285$

$pH=3.55$

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ( $HN_3$ ). So:

$HN_3~~H^+~+~N_3^-$

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

$Ka=\frac{[H^+][N_3^-]}{[HN_3]}$

For each mol of $H^+$ produced we will have 1 mol of $N_3^-$ . So, we can use "X" for the unknown values and replace in the Ka equation:

$Ka=\frac{X*X}{[HN_3]}$

Additionally, we have to keep in mind that $HN_3$ is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

$Ka=\frac{X*X}{0.004-X}$