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lora16 [44]
2 years ago
6

Okay help please will give brainlist

Chemistry
1 answer:
pashok25 [27]2 years ago
3 0

Answer:

b.

Explanation:

hope that helped you

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At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction. 2
xxMikexx [17]

Answer:

K = 3.37

Explanation:

2 NH₃(g) → N₂(g)  + 3H₂(g)

Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).

              2 NH₃(g)    →    N₂(g)  + 3H₂(g)

Initally       4moles             -            -

React        2moles           2m   +   3m

Eq             2 moles          2m        3m

We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)

The expression for K is:  ( [H₂]³ . [N₂] ) / [NH₃]²

We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)

K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²

K = 27/8 / 1 → 3.37

5 0
3 years ago
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A sample of pure oxalic acid (H2C2O4.2H2O) weighs 0.2000 g and requires 30.12 ml of KOH solution for
ValentinkaMS [17]

 The molarity  of KOH  is  0.1055 M

 <u><em> calculation</em></u>

Step  1: write  the  equation  for reaction between H₂C₂O₄.2H₂O  and KOH

H₂C₂O₄.2H₂O  + 2 KOH   →    K₂C₂O₄ +4 H₂O

step 2: find the moles  of H₂C₂O₄.2H₂O

moles = mass÷ molar  mass

from  periodic  table the  molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4)  + 2(18)=126 g/mol

 = 0.2000 g ÷ 126 g/mol =0.00159  moles


step 3: use the  mole  ratio  to  calculate the moles of KOH

H₂C₂O₄.2H₂O : KOH  is 1:2

therefore the  moles of KOH  =0.00159 x 2 = 0.00318  moles

step 4: find molarity of KOH

molarity = moles/volume in liters

volume in liters = 30.12/1000=0.03012 L

molarity  is therefore = 0.00318/0.03012 =0.1055 M

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The right answer for the question that is being asked and shown above is that: "a. Lower demand will prevent large increases in timber prices." it is important to reduce demands for timber because <span>a. Lower demand will prevent large increases in timber prices.</span>
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