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ELEN [110]
3 years ago
8

Is vinegar in water a true solution, suspension or colloid

Chemistry
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

Solution

Explanation:

Solution

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If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid
belka [17]

The balanced equation between the H_2SO_4 and NaHCO_3 is:

H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O

Formula of molarity is:

Molarity = \frac{Moles of solute}{Volume of solution in Liters}

Molarity = 6.0 M, Volume = 35 mL = 0.035 L

Substituting the values,

6 = \frac{Moles of solute}{0.035}

Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole

So, number of moles of H_2SO_4 is 0.21 mole.

From the balanced equation it is clear that for 1 mole of H_2SO_4, 2 moles of NaHCO_3 are required.

Hence, 0.21 mole of H_2SO_4  = 2\times 0.21 mole = 0.42 mole of NaHCO_3

Molar mass of NaHCO_3 = 84.007 g/mol

So, the mass of NaHCO_3 = 84.007 g/mol \times 0.42 mol = 35.283 g




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True or False: If you have 2 moles of water and 2 moles of carbon, you have the same number of molecules.
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Use the equation n2 + 3h2=2nh3 if 8.0g N2 react, how many grams of NH3 will be produced
Eduardwww [97]

the balanced equation for the formation of ammonia is

N₂ + 3H₂ ---> 2NH₃

molar ratio of N₂ to NH₃ is 1:2

mass of N₂ reacted is 8.0 g

therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol

according to the molar ratio,

1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant

therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃

therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g

a mass of 9.72 mol of NH₃ is formed

6 0
3 years ago
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