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Ostrovityanka [42]
3 years ago
7

The functional group for an ester is: RCHO RCOOH RCOOR ROH

Chemistry
1 answer:
Marizza181 [45]3 years ago
5 0

N,N-Phylumethaleneglycol-19

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A large room contains moist air at 308C, 102 kPa. The partial pressure of water vapor is 1.5 kPa. Determine
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What is the mass in grams of 0.280 mole sample of sodium hydroxide NaOH​
vivado [14]

Answer:

The mass of 0.280 mole sample of sodium hydroxide NaOH​ is 11.2 grams.

Explanation:

To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.

So you know:

  • Na: 23 g/mole
  • O: 16 g/mole
  • H: 1 g/mole

So, the molar mass of NaOH is:

NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole

Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?

mass=\frac{0.28 moles*40 grams}{1 mole}

mass= 11.2 grams

<u><em>The mass of 0.280 mole sample of sodium hydroxide NaOH​ is 11.2 grams.</em></u>

4 0
3 years ago
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
Fermium-253 has a half-life of 0.334 seconds. How long will it take, in seconds, for a 96 atom sample to decay to 3 atoms?​
iris [78.8K]

Answer:3.34 seconds

Explanation:

8 0
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