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Tanya [424]
3 years ago
5

Calculate the concentration in mol L^-1 of ammonium ions in a solution prepared by mixing 360 mL of 0.250 mol L^-1 ammonium sulf

ate solution with 675.0 mL of 1.20 mol L^-1 ammonium nitrate solution. Assume solutions are additive.
Chemistry
1 answer:
Mnenie [13.5K]3 years ago
8 0

In order to determine the concentration of ammonium ions in the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium nitrate, first calculate the amount of ammonium ions for each solution.<span>

<span>For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 = 0.18 mol NH4+
<span>For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3 = 0.81 mol NH4+

Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99 mol NH4+. To get the concentration, multiply this to the volume of solution which is assumed to be additive, such that:</span></span></span>

M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L sol’n

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It takes 26.8 mL of a 0.0700 M NaOH standard solution to neutralize a 250 mL sample of lactic acid (C3H6O3). What mass of lactic
Butoxors [25]
There is 1 OH- in 1 molecule of NaOH. 
Also, there is 1 H+ in 1 molecule of lactic acid.

So the reaction is simple.
so just equate the moles
moles of OH- in NaOH = moles of H+ in lactic acid
26.8 x 0.07 = 250 x Mole of lactic
Moles of lactic = 0.0075

so mass = 0.0075 x 90.8 =  0.681 g

4 0
3 years ago
Calculate the amount of carbon dioxid gas in 1.505x10^23 molecules of the gas.
Daniel [21]

Explanation:

  • We need to find the amount of carbon dioxide gas in 1.505\times 10^{23} molecules of the gas.
  • We know that, 1 mole weighs 44 gram of carbon dioxide which contains 6.022\times 10^{23} number of molecules. It means that, 6.022\times 10^{23} number of molecules present in 44 grams of carbon dioxide molecule. So,1.505\times 10^{23}  number of molecules present in :

=\dfrac{1.0505\times 10^{23}}{6.022\times 10^{23}}\times 44\\\\=7.675\ \text{grams}

  • Hence, 7.675 grams of carbon dioxide is present in 1.505\times 10^{23} molecules of the gas.
4 0
3 years ago
Consider the following reaction where Kc = 1.80×10-2 at 698 K:
Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

Qc=\frac{[C]^{c}*[D]^{d}  } {[A]^{a}*[B]^{b}}

In this case:

Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

  • [H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}=2.09*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}=4.14*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{0.280 moles}{1 Liter}= 0.280 \frac{moles}{liter}

So,

Qc=\frac{2.09*10^{-2} *4.14*10^{-2}  } {0.280^{2} }

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

8 0
3 years ago
Chemical ____ are used to represent compounds ​
cupoosta [38]

Answer:

Its formula :)

Explanation:

hope this helps <3

3 0
3 years ago
Read 2 more answers
YOU DO:
mihalych1998 [28]

Answer:

67.1%

Explanation:

Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:

<em>Moles Na₂CO₃ - 105.99g/mol-:</em>

6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.

As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:

0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃

And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):

0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.

And percent of NaHCO₃ in the sample is:

10.06g NaHCO₃ / 15g Sample * 100 =

<h3>67.1%</h3>
7 0
3 years ago
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