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Serga [27]
3 years ago
6

Popeye wants to make a dilute spinach solution for Sweetpea's bottle. How much 3.0 M spinach solution should he add to the 500.0

mL bottle if he wants the final solution to be 0.30 M?
Chemistry
1 answer:
Lynna [10]3 years ago
8 0

Answer : The volume of 3.0 M spinach solution added should be, 50 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of spinach solution.

M_2\text{ and }V_2 are the final molarity and volume of diluted spinach solution.

We are given:

M_1=3.0M\\V_1=?\\M_2=0.30M\\V_2=500.0mL

Now put all the given values in above equation, we get:

3.0M\times V_1=0.30M\times 500.0mL\\\\V_1=50mL

Hence, the volume of 3.0 M spinach solution added should be, 50 mL

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Hno3, a strong acid, is added to shift the ag2co3 equilibrium (equation 7.6) to the right. explain why the shift occurs.
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Answer 1)  When a strong acid like HNO_{3} reacts with Ag_{2} CO_{3} usually the equilibrium shifts to the right because

As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change."  So, in the given reaction when HNO_{3} reacts with Ag_{2} CO_{3}  it generates carbon dioxide and water as a by product, if we are adding HNO_{3} it will remove some of the CO_{3} molecule from the reaction mixture, which then tends to shift the equilibrium towards right.

Answer 2) The same would be observed in this case, if we replace HNO_{3} with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.

As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing Ag^{+} and Cl^{-} ions from the solution.
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3 years ago
Determine the volume of 12.3 grams of formaldehyde gas at STP?
postnew [5]
9.184 liters CH2O at STP

I think this is correct. Good luck
3 0
3 years ago
Read 2 more answers
Please answer both questions.
Deffense [45]

Answer:

1)Krypton

2)11H

Explanation:

electrons=protons

protons=atomic number

mass number=protons+neutrons

mass number is the superscript

atomic number is the subscript.

1)The answer is Krypton because its atomic number= number of protons=number of electrons is 36.

mass number is 46+36=82.

2)subscript=atomic number=number of protons=number of electrons

i. H = electrons=1

=neutrons=0

ii. Cl=electrons=17

=neutrons=35-17=18

iii. Na=electrons=11

=neutrons=23-11= 12

so the answer is Hydrogen because it has 1 electron and 0 neutron.

I hope this helps.

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3 years ago
Can depletion to the ozone cause changes to the biochemical cycles
GuDViN [60]
Hoi!! 

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3 years ago
Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat
Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values ​​of the atomic mass of each element that form the compounds:

  • N: 14 g/mol
  • H: 1 g/mol
  • O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

  • NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
  • O₂: 2*16 g/mol= 32 g/mol
  • NO: 14 g/mol + 16 g/mol= 30 g/mol
  • H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

  • NH₃: 4 moles
  • O₂: 5 moles
  • NO: 4 moles
  • H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

  • NH₃: 4 moles*17 g/mol= 68 g
  • O₂: 5 moles*32 g/mol= 160 g
  • NO: 4 moles*30 g/mol= 120 g
  • H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }

mass of O₂≅17.65 g

<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>

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