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Ratling [72]
3 years ago
5

What volume (in mLmL) of a 0.175 MHNO3MHNO3 solution is required to completely react with 44.2 mLmL of a 0.108 MNa2CO3MNa2CO3 so

lution according to the following balanced chemical equation? Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Answer:

54.5 mL

Explanation:

Let's consider the following balanced equation.

Na₂CO₃(aq) + 2 HNO₃(aq) → 2 NaNO₃(aq) + CO₂(g) + H₂O(l)

44.2 mL of 0.108 M Na₂CO₃ solution react. The moles of Na₂CO₃ that react are:

0.0442 L × 0.108 mol/L = 4.77 × 10⁻³ mol

The molar ratio of Na₂CO₃ to HNO₃ is 1:2. The moles of HNO₃ that reacted are  2/1 × 4.77 × 10⁻³ mol = 9.54 × 10⁻³ mol

9.54 × 10⁻³ moles of HNO₃ are in a 0.175 M solution. The required volume of HNO₃ is:

9.54 × 10⁻³ mol × (1 L/0.175 mol) × (1000 mL/1 L) = 54.5 mL

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A covalent bond between an oxygen atom and a hydrogen atom is best described as _____.
Igoryamba
The answer is most likely nonmetals. :)
7 0
3 years ago
Given the following reaction: NH4SH (s) <--> NH3 (g) + H2S (g) If we start
almond37 [142]

Answer:

D. 0.3 M

Explanation:

                                              NH4SH (s)      <-->            NH3 (g) + H2S (g)

Initial concentration              0.085mol/0.25L             0                 0

Change in concentration     -0.2M                               +0.2 M        +0.2M

Equilibrium               0.035mol/0.25 L=0.14M             0.2M           0.2M

concentration

Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M

K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M

4 0
3 years ago
A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

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Answer:

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Explanation:

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