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Marrrta [24]
3 years ago
13

You have 5 kg of water, and add 5 moles of NaCL to it. what is the molarity of the solution?

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
4 0
As Molarity is given by,
                                     Molarity = moles / volume     ------(1)
Data Given:
                   mass of water = 5 Kg = 5000 g
                                moles = 5 moles
Solution:
As we know that volume is related to mass as,
                                     Volume = mass / density
Putting values,
                                     Volume = 5000 g / 1 g/cm³
                                               V = 5000 cm³
Now, Putting value of volume and moles in eq. 1,
                                               M = 5 mol / 5000 cm³
                                               M = 0.0001 mol/cm³
or,
                                               M = 1 mol/dm³
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Wich reason is the best for choosing the ruler to measure the dimensionof tbe wooden block
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NH₄NO₃ → N₂O + 2H₂O When 45.70 g of NH₄NO₃ decomposes, what mass of each product is formed?
Anna007 [38]

Answer: 25.13 g of N_2O  and 20.56 g of H_2O will be produced from 45.70 g of NH_4NO_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles

The balanced chemical equation is:

NH_4NO_3\rightarrow N_2O+2H_2O  

According to stoichiometry :

1 mole of NH_4NO_3 produce = 1 mole of N_2O

Thus 0.571 moles of NH_4NO_3 will require=\frac{1}{1}\times 0.571=0.571moles  of N_2O  

Mass of N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g

1 mole of NH_4NO_3 produce = 2 moles of H_2O

Thus 0.571 moles of NH_4NO_3 will require=\frac{2}{1}\times 0.571=1.142moles  of H_2O  

Mass of H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g

Thus 25.13 g of N_2O  and 20.56 g of H_2O will be produced from 45.70 g of NH_4NO_3

5 0
3 years ago
How many grams of oxygen are required to burn 13.5 g of acetylene
Vinil7 [7]

Answer:

41.54 grams of oxygen are required to burn 13.5 g of acetylene

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • C₂H₂: 2 moles
  • O₂: 5 moles
  • CO₂: 4 moles
  • H₂O: 2 moles

Being the molar mass of the compounds:

  • C₂H₂: 26 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • C₂H₂: 2 moles* 26 g/mole= 52 grams
  • O₂: 5 moles* 32 g/mole= 160 grams
  • CO₂: 4 moles* 44 g/mole= 176 grams
  • H₂O: 2 moles* 18 g/mole= 36 grams

You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?

mass of oxygen=\frac{13.5 grams of acetylene*160 grams of oxygen}{52 grams of acetylene}

mass of oxygen= 41.54 grams

<u><em>41.54 grams of oxygen are required to burn 13.5 g of acetylene</em></u>

<u><em></em></u>

8 0
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myrzilka [38]
laboratory stock solution is 1.50 M NaOH. Calcul. Show transcribed image text A laboratory stock solution is 1.50 M NaOH. Calculate the volume of this stock solution that would be needed to prepare 300. ml
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